Q52 / 120 · JEE Advanced 2006 · Mathematics
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
- A0
- B1
- C$\sqrt{2}$
- Correct answerD$2 \sqrt{2}$
Solution
Let the equation of plane be,
$$
a(x-1)+b(y+2)+c(z-1)=0
$$
which is perpendicular to $2 x-2 y+z=0$ and $x-y+2 z=4$
$$
\begin{array}{llrl}
\Rightarrow & 2 a-2 b+c & =0 \text { and } a-b+2 c=0 \\
\text { or } & \frac{a}{-2} 1 & =\frac{b}{2}=\frac{c}{-2} \\
\Rightarrow & -12 & 1 & -1 \\
\Rightarrow & \frac{a}{-3} & =\frac{b}{-3}=\frac{c}{0} \\
& \frac{a}{1}=\frac{b}{1}=\frac{c}{0}
\end{array}
$$
So, the equation of plane is,
$x-1+y+2=0$ or $x+y+1=0$, its distance from the point $(1,2,2)$ is $\frac{|1+2+1|}{\sqrt{2}}=2 \sqrt{2}$.
Hence, the correct answer is option: 'D'.