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If the image of the point P(1, −2, 3) in the plane, 2x+3y−4z+22=0 measured parallel to the line, \frac{x}{1}= \frac{y}{4}= \frac{z}{5}    is Q, then PQ is equal to :
Option: 1 2\sqrt{42}

Option: 5 \sqrt{42}

Option: 9 6\sqrt{5}

Option: 13 3\sqrt{5}
 

 

Intersection of line and plane -

Let the line

\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c} plane

a_{1}x+b_{1}y+c_{1}z+d=0 intersect at P

to find P assume general point on line as \left ( x_{1}+\lambda a_{1}y_{1} +\lambda b_{1}z_{1}+\lambda c_{1}\right )

now put it in plane to find \lambda,

a_{1}\left ( x_{1}+\lambda a \right )+b_{1}\left ( y_{1}+\lambda b \right )+c_{1}\left ( z_{1}+\lambda c \right )+d=0

-

 

 

Distance formula -

The distance between two points A(x_{1},y_{1},z_{1})\, and \, B(x_{2},y_{2},z_{2}) is =\sqrt{\left ( x_{2}-x_{1} \right )^{2}+{\left ( y_{2}-y_{1} \right )^{2}}+{\left ( z_{2}-z_{1} \right )^{2}}}

 

- wherein

\vec{OA}= x_{1}\hat{i}+y_{1}\hat{j}+z_{1}\hat{k}

\vec{OB}= x_{2}\hat{i}+y_{2}\hat{j}+z_{2}\hat{k}

\underset{AB}{\rightarrow}    = \underset{OB}{\rightarrow} - \underset{AB}{\rightarrow}

\underset{AB}{\rightarrow}  = \left ( x_{2}-x_{1} \right )\hat{i}+\left ( y_{2}-y_{1} \right )\hat{j}+\left ( z_{2}-z_{1} \right )\hat{k}

AB= \left | \underset{AB}{\rightarrow} \right |

AB= \sqrt{\left ( x_{2}-x_{1} \right )^{2}+\left ( y_{2}-y_{1} \right )^{2}+\left ( z_{2}-z_{1} \right )^{2}}

 

 

 

 

PQ=2PM

Now, for finding M

\frac{x}{1}=\frac{y}{4}=\frac{z}{5}=k(say)

\Rightarrow x=k,y=4k,z=5k

Putting in equation of plane,

k=\frac{11}{3}

So, M=\left (\frac{11}{3} ,\frac{44}{3},\frac{55}{3} \right )

\therefore PM=\sqrt{\frac{64}{9}+\frac{2500}{9}+\frac{2116}{9}}= 2\sqrt{130}

\Rightarrow PQ=4\sqrt{130}

 

 

 

 

 

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vishal kumar

The line of intersection of the planes \small \underset{r}{\rightarrow} .\left ( 3\hat{i} -\hat{j} +\hat{k}\right )= 1 and \small \underset{r}{\rightarrow} .\left ( \hat{i} +4 \hat{j} -2\hat{k}\right ) =2, is:
Option: 1 -2 \hat{i}+7 \hat{j}+13 \hat{k}
Option: 2 2 \hat{i}+7 \hat{j}-13 \hat{k}
Option: 3 -2 \hat{i}-7 \hat{j}+13 \hat{k}
Option: 4 -2 \hat{i}+7 \hat{j}+13 \hat{k}
 

As we have learned

Equation of line as intersection of two planes -

Let the two intersecting planes be

ax+by+cz+d= 0 and 

a_{1}x+b_{1}y+c_{1}z+d_{1}= 0

then the parallel vector of line formed their intersection can be obtained by

\begin{vmatrix} \hat{i} &\hat{j} & \hat{k}\\ a&b &c \\ a_{1} & b_{1} & c_{1} \end{vmatrix}= A\hat{i}+B\hat{j}+C\hat{k}(assumed)

and points can be obtained by putting z= 0 and solving

ax+by+d= 0 and 

a_{1}x+b_{1}y+d_{1}= 0 say \alpha ,\beta

Now the equation will be

\frac{x-\alpha }{A}=\frac{y-\beta }{B}=\frac{z-0 }{C}

 

-

 

 3x-y + z = 1 \\ and \: \: x + 4y - 2z = 2

putting z = 0 

3x-y = 1    and   x + 4y = 2 

\Rightarrow 13 x = 6 \Rightarrow x = 6/13 \\ y = 5/3

vector \: \: along \: \: line = \begin{vmatrix} \hat i & \hat j &\hat k \\ 3 & -1 & 1\\ 1 & 4 &-2 \end{vmatrix} = -2 \hat i + 7 \hat j +13 \hat k

 

 

 

 

 

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The shortest distance between the lines \frac{x-3}{3}=\frac{y-8}{-1}=\frac{z-3}{1} and \frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-6}{4} is :
 
Option: 1 2\sqrt{30}
Option: 2 \frac{7}{2}\sqrt{30}
Option: 3 3
Option: 4 3\sqrt{30}
 

 

 

Shortest Distance between Two Lines -

Distance between two skew lines

If L1 and L2  are two skew lines, then there is one and only one line perpendicular to each of lines L1 and L2 which is known as the line of shortest distance.

Vector form

\\\mathrm{\;\;\;L_1:\;\;\;\;}\vec{\mathbf r} &=\vec{\mathbf r}_{0}+\lambda \vec{\mathbf b} \\\\\mathrm{\;\;\;L_2:\;\;\;\;}\vec{\mathbf r} &=\vec{\mathbf r'}_{0}+\mu \vec{\mathbf b'}

If \overrightarrow{PQ} is the shortest distance vector between L1 and L2, then it being perpendicular to both \vec{\mathbf b} and \vec{\mathbf b'}, therefore, the unit vector \hat{\mathbf n} along \overrightarrow{PQ} would be 

\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\hat{\mathbf n}=\frac{\vec {\mathbf b}\times \vec{\mathbf b'}}{\left | \vec{\mathbf b}\times \vec{\mathbf b'} \right |}\\\mathrm{Then,\;\;\;\;\;\;\;\;\;\;\;\;}\overrightarrow{PQ}=\mathit{d}\hat{\mathbf n}

where "d" is the magnitude of the shortest distance vector. Let θ be the angle between \overrightarrow{ST} and \overrightarrow{PQ}.

Then  

\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;PQ=ST\left |\cos\theta \right |}\\\\\text{but,}\;\;\;\;\;\;\;\;\;\;\;\;\;\cos \theta=\left|\frac{\overrightarrow{\mathrm{PQ}} \cdot \overrightarrow{\mathrm{ST}}}{|\overrightarrow{\mathrm{PQ}}||\overrightarrow{\mathrm{ST}}|}\right|\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}=\left|\frac{d \hat{n} \cdot\left(\vec{r'}_{0}-\vec{r}_{0}\right)}{d \;\mathrm{ST}}\right| \quad\left(\text { since } \overrightarrow{\mathrm{\;ST}}=\vec{r'}_{0}-\vec{r}_{0}\right)\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}=\left|\frac{\left(\vec{b} \times \vec{b'}\right) \cdot\left(\vec{r'}_{0}-\vec{r}_{0}\right)}{\mathrm{ST}\left|\vec{b} \times \vec{b'}\right|}\right|

Hence, the required shortest distance is

\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}d=\mathrm{PQ}=\mathrm{ST}|\cos \theta|\\\\\text{or}\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\mathbf d=\left|\frac{\left(\vec{\mathbf b} \times \vec{\mathbf b'}\right) \cdot\left(\vec{\mathbf r'}_{0} - \vec{\mathbf r}_{0}\right)}{\left|\vec{\mathbf b} \times \vec{\mathbf b'}\right|}\right|

-

 

 

\\\overrightarrow{\mathrm{AB}}=6 \hat{i}+15 \hat{j}+3 \hat{k}\\\vec p=3\hat i-\hat j+\hat k\\ \vec q=-3\hat i+2\hat j+4\hat k\\\vec p\times \vec q=\begin{vmatrix} \hat i &\hat j &\hat k \\ 3 &-1 &1 \\ -3 & 2 &4 \end{vmatrix}=\hat i(-4-2)-\hat j(12+3)+\hat k(6-3)\\\Rightarrow -6\hat i-15\hat j+3\hat k\\\text{shortest dist}=\frac{| \overrightarrow{\mathrm{AB}} \cdot(\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}})}{|\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}|}

=3\sqrt{30}

Correct Option (4)

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Kuldeep Maurya

Let P be a plane passing through the points (2,1,0), (4,1,1) and (5,0,1)and R be any point (2,1,6).Then the image of R in the plane P is :
Option: 1 (6,5,2)
Option: 2 (6,5,-2)
Option: 3 (4,3,2)
Option: 4 (3,4,-2)
 

 

 

Equation of a plane passing through three non collinear point -

 

Cartesian Form

Let (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3) be the coordinates of points A, B and C respectively. 

Let P(x, y z) be any point on the plane.

Then, the vectors \overrightarrow{PA},\;\overrightarrow{BA} and \overrightarrow{CA} are coplanar.
\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\left [ \overrightarrow{\mathbf{PA}}\;\;\overrightarrow{\mathbf{BA}}\;\;\overrightarrow{\mathbf{CA}} \right ]=0\\\\\mathrm{\;\;\;}{\begin{vmatrix} \mathbf{x-x_1 }&\mathbf{y-y_1} &\mathbf{z-z_1} \\ \mathbf{x_2-x_1} &\mathbf{y_2-y_1} &\mathbf{z_2-z_1} \\ \mathbf{x_3-x_1} &\mathbf{y_3-y_1} &\mathbf{z_3-z_1} \end{vmatrix}=0}\\\\\text{which is required equation of the plane .}

-

 

 

Image of a Point in the Plane -

The image of the point P(x1, y1, z1) in the plane ax + by + cz + d = 0 is  Q(x3, y3, z3) then coordinates of point Q is given by

\mathbf{\frac{x_3-x_1}{a}=\frac{y_3-y_1}{b}=\frac{z_3-z_1}{c}=-\frac{2\left ( ax_1+by_1+cz_1+d \right )}{{a^2+b^2+c^2}}} 

-

Points of plane P ( 2,1,0), (4,1,1)  and ( 5,0,1)  and point R ( 2,1,6) . 

Then the image of R in the plane P is: 

\begin{vmatrix} x-2 &y-1 &z \\ 2&0 &1 \\ 3& -1& 1 \end{vmatrix} = (x-2)[0+1]-(y-1)[2-3]+ z[-2] \\ \\ \Rightarrow x-2+y-1-2z=0\\ \text{plane} , x+y-2z=3

\\\Rightarrow \frac{x-2}{1}=\frac{y-1}{1}=\frac{z-6}{-2}=\frac{-2(2+1-12-3)}{6} \\\Rightarrow(x, y, z)=(6,5,-2)

Correct option (2)

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A vector \vec{a}=\alpha \hat{i}+2\hat{j}+\beta \hat{k}\; (\alpha ,\beta \epsilon R) lies in the plane of the vectors, \vec{b}= \hat{i}+\hat{j} and \vec{c}= \hat{i}-\hat{j}+4\hat{k}. If \vec{a} bisects the angle between \vec{b} and \vec{c}, then:
Option: 1 \vec{a}\cdot \hat{i}+3=0
Option: 2 \vec{a}\cdot \hat{k}+4=0
Option: 3 \vec{a}\cdot \hat{i}+1=0
Option: 4  \vec{a}\cdot \hat{k}+2=0
 

 

 

Equation of The Plane Bisecting the Angle Between Two Planes -

-

angle bisector can be \vec{a}=\vec{a}=\lambda(\hat{b}+\hat{c}) \text { or } \vec{a}=\mu(\hat{b}-\hat{c})

\\\overrightarrow{\mathrm{a}}=\lambda\left(\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{\sqrt{2}}+\frac{\hat{\mathrm{i}}-\hat{\mathrm{j}}+4 \hat{\mathrm{k}}}{3 \sqrt{2}}\right)=\frac{\lambda}{3 \sqrt{2}}[3 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+\hat{\mathrm{i}}-\hat{\mathrm{j}}+4 \hat{\mathrm{k}}]\\=\frac{\lambda}{3 \sqrt{2}}[4 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}]

Compare it with \overrightarrow{\mathrm{a}}=\alpha \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\beta \hat{\mathrm{k}}

\\ {\frac{2 \lambda}{3 \sqrt{2}}=2 \Rightarrow \lambda=3 \sqrt{2}} \\ {\overrightarrow{\mathrm{a}}=4 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}} \\ {\text { Not in option so now consider }} \overrightarrow{\mathrm{a}}=\mu\left(\frac{\hat{i}+\hat{j}}{\sqrt{2}}-\frac{\hat{i}-\hat{j}+4 \hat{k}}{3 \sqrt{2}}\right)

\begin{aligned} \overrightarrow{\mathrm{a}}=& \frac{\mu}{3 \sqrt{2}}(3 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-\hat{\mathrm{i}}+\hat{\mathrm{j}}-4 \hat{\mathrm{k}}) \\=& \frac{\mu}{3 \sqrt{2}}(2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}-4 \hat{\mathrm{k}}) \end{aligned}

Compare it with \overrightarrow{\mathrm{a}}=\alpha \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\beta \hat{\mathrm{k}}

\\ {\frac{4 \mu}{3 \sqrt{2}}=2 \Rightarrow \mu=\frac{3 \sqrt{2}}{2}} \\ {\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}} \\ {\overrightarrow{\mathrm{a}} \cdot \hat{\mathrm{k}}+2=0} \\ {-2+2=0}

Correct Option (4)

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Ritika Jonwal

The mirror image of the point (1,2,3) in a plane is \left ( \frac{-7}{3},-\frac{4}{3},-\frac{1}{3} \right ). which of the following points lies on this plane ?
Option: 1 (1,-1,1)
Option: 2 (-1,-1,1)
Option: 3 (1,1,1)
Option: 4 (-1,-1,-1)
 

 

 

Image of a Point in the Plane -

The image of the point P(x1, y1, z1) in the plane ax + by + cz + d = 0 is  Q(x3, y3, z3) then coordinates of point Q is given by

\mathbf{\frac{x_3-x_1}{a}=\frac{y_3-y_1}{b}=\frac{z_3-z_1}{c}=-\frac{2\left ( ax_1+by_1+cz_1+d \right )}{{a^2+b^2+c^2}}} 

Proof:

Let point Q(x3, y3, z3) is the image of the point P(x1, y1, z1) in the plane π : ax + by + cz + d = 0.

Let line PQ meets the plane ax+by+cz+d=0 at point R, direction ratio of normal to plane π are (a, b, c), since, PQ perpendicular to plane π.

So direction ratio of PQ are a, b, c 

\\\Rightarrow \text{equation of line PQ is}\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}=r\;\;\;\;\text{(say)}\\\\\text { any point on line } P Q \text { may be taken as }\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\left ( ar+x_1,br+y_1,cr+z_1 \right )\\\mathrm{Let\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}Q\equiv\left ( ar+x_1,br+y_1,cr+z_1 \right )\\\text {since, } R \text { is the middle point of } P Q\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}R\equiv\left ( \frac{x_1+ar+x_1}{2},\frac{y_1+br+y_1}{2},\frac{z_1+cr+z_1 }{2}\right )\\\mathrm{\Rightarrow \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}R\equiv\left ( x_1+\frac{ar}{2},y_1+\frac{br}{2},z_1+\frac{cr }{2}\right )

Since, point R lies in the plane  π, we get

\\\mathrm{\Rightarrow \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}R\equiv\left ( x_1+\frac{ar}{2},y_1+\frac{br}{2},z_1+\frac{cr }{2}\right )\\\mathrm{\Rightarrow \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}a\left ( x_1+\frac{ar}{2}\right )+b\left ( y_1+\frac{br}{2} \right )+ c\left ( z_1+\frac{cr }{2} \right )+d=0\\\\\Rightarrow \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left ( a^2+b^2+c^2 \right )\frac{r}{2}=-\left ( ax_1+by_1+cz_1+d \right )\\\\\Rightarrow \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{r}=\frac{-2\left ( ax_1+by_1+cz_1+d \right )}{a^2+b^2+c^2}

In the similar method we can also find the coordinates of point R.

-

 

 

 

P(1,2,3) and Q\left(\frac{-7}{3}, \frac{-4}{3}, \frac{-1}{3}\right)

Direction ratio of normal to the plane \left(1-\frac{-7}{3}, 2-\frac{-4}{3}, 3-\frac{-1}{3}\right)

we get \frac{10}{3}, \frac{10}{3}, \frac{10}{3}

and Midpoint of PQ is \left(\frac{-2}{3}, \frac{1}{3}, \frac{4}{3}\right)

equation of plane x + y + z = 1

Correct Option (1)

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vishal kumar

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If the foot of the perpendicular drawn from the point (1,0,3) on line passing through (\alpha ,7,1) is \left ( \frac{5}{3},\frac{7}{3},\frac{17}{3} \right ), then \alphais equal to _______.
Option: 1 2
Option: 2 4
Option: 3 8
Option: 4 10
 

 

 

Angle Between Two Lines -

Vector Form
Condition for Perpendicularity

The lines are perpendicular then cos? = 90o

\\\teaxt{i.e.}\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\vec{\mathbf b}\cdot\vec{\mathbf b}'=0\;\;\;\;\;\;\;\;\;\;\;\;\; \left [\because\;\; \cos90^\circ=0 \right ]\\\\\Rightarrow\;\;\;\;a_1a_2+b_1b_2+c_1c_2=0


Condition for parallelism

The lines are parallel then \vec{\mathbf b}=\lambda\vec{\mathbf b}' for some scalar λ.

\Rightarrow \;\;\;\;\;\;\;\;\;\;\;\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}

-

Since PQ is perpendicular to L, therefore

\\\left ( 1-\frac{5}{3} \right )\left ( \alpha-\frac{5}{3} \right )+\left ( -\frac{7}{3} \right )\left ( 7-\frac{7}{3} \right )+\left ( 3-\frac{17}{3} \right )\left ( 1-\frac{17}{3} \right )=0\\\alpha=4

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Ritika Jonwal

Let the foot of perpendicular from a point \mathrm{P}(1,2,-1) to the straight line L: \frac{x}{1}=\frac{y}{0}=\frac{z}{-1}$ be $N. Let a line be drawn from P parallel to the plane \mathrm{x}+\mathrm{y}+2 \mathrm{z}=0 which meets \mathrm{L}  at point \mathrm{Q}. If \alpha is the acute angle between the lines P N$ and $P Q, then \cos \alpha is equal to__________.
Option: 1 \frac{1}{\sqrt{5}}
Option: 2 \frac{\sqrt{3}}{2}
Option: 3 \frac{1}{\sqrt{3}}
Option: 4 \frac{1}{2 \sqrt{3}}


Let \, N\, be\, \frac{x}{1}= \frac{y}{0}= \frac{z}{-1}= \lambda \Rightarrow x= \lambda ,y= 0,z= -\lambda
N\left ( \lambda ,0,- \lambda \right )
Now\, PN\perp L\Rightarrow \overrightarrow{PN}\cdot \left ( i-k \right )= 0
\Rightarrow \left ( \left ( \lambda -1 \right )i-2j+\left ( -\lambda+1 \right )k \right )\cdot \left ( i-k \right )= 0
\Rightarrow \left ( \lambda -1 \right )+\lambda -1= 0\Rightarrow \lambda = 1
\Rightarrow N\left ( 1,0,-1 \right )



Let\, Q\, be\, \left ( p,0,-p \right )
\therefore \overrightarrow{PQ}\perp \left ( normal\, to \, plane \right )
\Rightarrow \overrightarrow{PQ}\cdot \vec{n}= 0
\Rightarrow \left [ \left ( p-1 \right ) i+\left ( -2 \right )j+\left ( -p+1 \right )k\right ]\cdot \left ( i+j+2k \right )= 0
\Rightarrow p= -1
\Rightarrow Q\left ( -1,0,1 \right )


\therefore \overrightarrow{PN}= 2j\, and\, \overrightarrow{PQ}= -2i-2j+2k
Angle\, between\, these= \cos^{-1}\left | \frac{\overrightarrow{PN}\cdot \overrightarrow{PQ}}{\left | \overrightarrow{PN} \right |\left | \overrightarrow{PQ} \right |} \right |
                                           = \cos^{-1}\left | \frac{4}{2\cdot 2\sqrt{3}} \right |= \cos^{-1}\left ( \frac{1}{\sqrt{3}} \right )

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Let the acute angle bisector of the two planes x-2 y-2 z+1=0$ and $2 x-3 y-6 z+1=0 be the plane P. Then which of the following points lies on P ?
Option: 1 (0,2,-4)
Option: 2 \left(-2,0,-\frac{1}{2}\right)
Option: 3 (4,0,-2)
Option: 4 \left(3,1,-\frac{1}{2}\right)

The angle bisectors are

\frac{x-2y-2z+1}{\sqrt{1^{1}+2^{2}+2^{2}}}= \pm \frac{2x-3y-6z+1}{\sqrt{2^{2}+3^{2}+6^{2}}}

\Rightarrow 7\left ( x-2y-2z+1 \right )= \pm 3\left ( 2x-3y-6z+1 \right )\Rightarrow7x-14y-14z+7= \pm \left ( 6x-9y-18z+3 \right )

\Rightarrow 13x-23y-32y+10= 0----\left ( 1 \right )
    OR     x-5y+4z+4= 0---\left ( 2 \right )

Take a point (-1,0,0) on plane 1 and find its distance from both angle bisectors
d_{1}= \left | \frac{-13+10}{\sqrt{13^{2}+23^{2}+32^{2}}} \right |= \frac{3}{\sqrt{1722}}
d_{2}= \left |\frac{-1}{\sqrt{1^{2}+5^{2}+4^{2}}}\right |= \frac{1}{\sqrt{42}}= \frac{3}{\sqrt{368}}

Clearly d_{1}< d_{2}

So plane in equation (1) is acute angle Bisector
13x-23y-32z+10= 0
Out of the given option, \left ( -2,0,\frac{-1}{2} \right ) satisfies this plane

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Kuldeep Maurya

The distance of line 3 y-2 z-1=0=3 x-z+4 from the point (2,-1,6) is :
Option: 1 2 \sqrt{5}
Option: 2 2 \sqrt{6}
Option: 3 \sqrt{26}
Option: 4 4 \sqrt{2}

Let PQ be perpendicular to the given line

Now Q lies on both the planes,

So 3a-2c-1= 0---\left ( i \right ) and
     3a-c+4= 0---\left ( ii \right )

Also\: PQ\perp line\Rightarrow PQ\perp \left ( \vec{n_{1} }\times\vec{n_{2}} \right )
\Rightarrow \vec{PQ}\cdot \left ( \vec{n_{1}} \times \vec{n_{2}}\right )= 0
\Rightarrow\left [ \vec{PQ}\; \; \vec{n_{1}} \; \; \vec{n_{2}} \right ] = 0
\Rightarrow \begin{vmatrix} a-2&b+1 &c-6 \\ 3& 0 & -1\\ 0&3 &-2 \end{vmatrix}= 0
\Rightarrow 3\left ( a-2 \right )-\left ( b+1 \right )\left ( -6 \right )+\left ( c-6 \right )9= 0
\Rightarrow a-2+2b+2+3c-18= 0
\Rightarrow a+2b+3c= 18---\left ( iii \right )

Solving (i),(ii) & (iii)

a= 0,b= 3,c= 4\Rightarrow Q\left ( 0,3,4 \right )
PQ= 2\sqrt{6}

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