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Q.
A plane passes through $(1,-2,1)$ and is perpendicular to two planes $2 x-2 y+z=0$ and $x-y+2 z=4$, then the distance of the plane from the point $(1,2,2)$ is.
Three Dimensional Geometry
Solution:
D.R. of the normal to the plane that that is perpendicular to the plane
$2 x-2 y+z=0$ and $x-y+$ $2 z =4$ is given by
$\begin{vmatrix}\hat{ i } & \hat{ j } & \hat{ k } \\ 2 & -2 & 1 \\ 1 & -1 & 2\end{vmatrix}=-3 \hat{ i }-3 \hat{ j }$ or $\hat{ i }+\hat{ j }$
So the required plane passing through $(1,-2,1)$ is
$(x-1)+(y+2)=0$ or $x+y+1=0$
Hence the distance of $(1,2,2)$ form this plane is
$d =\frac{4}{\sqrt{2}}=2 \sqrt{2}$ units