A plane which is perpendicular to two planes $2 x-2 y+z=0$ and $x-y+2 z=4$, passes through $(1,-2,1)$. The…
A plane which is perpendicular to two planes $2 x-2 y+z=0$ and $x-y+2 z=4$, passes through $(1,-2,1)$. The distance of the plane from the point $(1,2,2)$ is
0 units
1 units
$\sqrt{2}$ units
$2 \sqrt{2}$ units
Solution
The equation of a plane passing through $(1,-2,1)$ is
$\mathrm{a}(x-1)+\mathrm{b}(y+2)+\mathrm{c}(\mathrm{z}-1)=0...(i)$
$\begin{aligned}
& \text { Plane (i) is perpendicular to planes } \\
& 2 x-2 y+z=0 \text { and } x-y+2 z=4 . \\
& \therefore \quad \begin{array}{l}
2 a-2 b+c=0, \text { and } ...(i)\\
a-b+2 c=0 ...(ii)\\
\text { Solving (ii) and (iii), we get } \\
a=-3, b=-3, c=0
\end{array}
\end{aligned}$
Substituting the values of $a, b, c$ in equation (i), we get
$x+y+1=0$
$\therefore \quad$ The distance of this plane from $(1,2,2)$ is
$d=\left|\frac{1+2+1}{\sqrt{1+1}}\right|=2 \sqrt{2}$