The equation of the plane passing through $(-2,2,2)$ and $(2,-2,-2)$ and perpendicular to the plane $9 x-13…
The 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
$5 x-3 y+2 z=12$
$5 x+3 y+2 z=0$
$5 x+3 y-2 z+8=0$
$5 x-3 y+2 z+12=0$
Solution
Equation of plane passing through $(-2,2,2)$ is $a(x+2)+b$
$(\mathrm{y}-2)+\mathrm{c}(\mathrm{z}-2)=0$
Since this plane also passes through $(2,-2,-2)$, we get
$4 \mathrm{a}-4 \mathrm{~b}-4 \mathrm{c}=0 \Rightarrow \mathrm{a}-\mathrm{b}-\mathrm{c}=0$
Normal of the plane is parallel to
$9 x-13 y-3 z=0 \Rightarrow 9 a-13 b-3 c=0$
Solving (1) and (2), we write
$\frac{\mathrm{a}}{\left|\begin{array}{cc}
-1 & -1 \\
-13 & -3
\end{array}\right|}=\frac{\mathrm{b}}{\left|\begin{array}{cc}
1 & -1 \\
0 & -3
\end{array}\right|}=\frac{\mathrm{c}}{\left|\begin{array}{cc}
1 & -1 \\
9 & -13
\end{array}\right|}$
$\therefore \frac{\mathrm{a}}{-10}=\frac{-\mathrm{b}}{6}=\frac{\mathrm{c}}{-4} \Rightarrow \frac{\mathrm{a}}{5}=\frac{\mathrm{b}}{3}=\frac{\mathrm{c}}{2}$
Hence equation of required plane is
$5(x+2)+3(y-2)+2(z-2)=0 \text { i.e. } 5 x+3 y+2 z=0$