Q69 / 150 · MHT CET 2009 · Mathematics
Equation of the plane passing through $(-2,2,2)$ and $(2,-2,-2)$ and perpendicular to the plane $9 x-13 y-3 z=0$ is
- Correct answerA$5 x+3 y+2 z=0$
- B$5 x-3 y+2 z=0$
- C$5 x-3 y-2 z=0$
- D$5 x+3 y-2 z=0$
Solution
Any plane passing through $(-2,2,2)$ is $A(x+2)+B(y-2)+C(z-2)=0$
$\because$ It passes through $(2,-2,-2)$ $\Rightarrow \quad 4 A-4 B-4 C=0$
It is parallel to $9 x-13 y-3 z=0$ $\begin{array}{lll}\therefore & 9 A-13 B-3 C=0 & \ldots \text { (ii) }\end{array}$
Solving Eqs. (i) and (ii),
$$
\begin{aligned}
& \frac{A}{12-52}=\frac{B}{-36+12}=\frac{C}{-52+36} \\
\Rightarrow \quad \frac{A}{-40}=\frac{B}{-24}=\frac{C}{-16}
\end{aligned}
$$
$\therefore$ Required equation of plane is $-40(x+2)-24(y-2)-16(z-2)=0$
$\begin{array}{lr}\Rightarrow & -40 x-80-24 y+48-16 z+32=0 \\ \Rightarrow & 40 x+24 y+16 z=0 \\ \Rightarrow & 5 x+3 y+2 z=0\end{array}$
Hence, the correct answer is option: 'A'.