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Let the line $\frac{x-2}{3}= \frac{y-1}{-5}= \frac{z+2}{2}$ lie in the plane

$x+3y-\alpha z+\beta = 0$ .Then $\left ( \alpha ,\beta \right )$ equals to

Option 1)
$(-6,7)$

Option 2)
$(5,-15)$

Option 3)
$(-5,5)$

Option 4)
$(6,-17)$

Answers (1)

best_answer

As we learnt in

Condition for line to be lie in plane -

$\vec{b}\cdot \vec{n}= 0$ and $\vec{a}\cdot \vec{n}= d$ or

$a_{1}a+b_{1}b+c_{1}c= 0$ and $a_{1}x_{1}+b_{1}y_{1}+c_{1}z_{1}+d= 0$

-

$\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$

(2,1,-2) lies in $x+3y-\alpha z+ \beta=0$

$2+3+2\alpha + \beta=0$

$2\alpha +\beta =-5$

Also, $(\hat{i}+3\hat{j} -\hat{k}). (3\hat{i}-5\hat{j}+2\hat{k})=0$

Since line 1 isperpendicular to normal plane, 3-15-2k=0

$\Rightarrow$ k= -6

Option 1)

$(-6,7)$

Correct answer

Option 2)

$(5,-15)$

Incorrect answer

Option 3)

$(-5,5)$

Incorrect answer

Option 4)

$(6,-17)$

Incorrect answer

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