# The distance of the point (1, −2, 4) from the plane passing through the point (1, 2, 2) and perpendicular to the planes x − y + 2z = 3 and  2x −2y + z + 12=0, is : Option 1) Option 2) Option 3) Option 4)

S Sabhrant Ambastha

As we learnt in

Distance of a point from plane (Cartesian form) -

The length of perpendicular from $P(x_{1},y_{1},z_{1})$ to the plane

$ax+by+cz+d= 0$ is given by  $\frac{\left [ ax_{1}+by_{1} +cz_{1}+d\right ]}{\left | \sqrt{a^{2}+b^{2}+c^{2}} \right |}$

-

Normal vector of plane is $\begin{vmatrix} \hat{i} &\hat{j} & \hat{k}\\ 1 & -1 & 2\\ 2& -2& 1 \end{vmatrix}=3\hat{i}+3\hat{j}$

So equation of plane is

$3x+3y=C$

passes through (1,2,2)

3+6=C

3x+3y=9    =>    x+y=3

Distance $=\left | \frac{1-2-3}{\sqrt{2}} \right |=\frac{4}{\sqrt{2}}=2\sqrt{2}$

Option 1)

This option is correct

Option 2)

This option is incorrect

Option 3)

This option is incorrect

Option 4)

This option is incorrect

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