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

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
\(
5(x+2)+3(y-2)+2(z-2)=0 \text { i.e. } 5 x\) \(+~3 y+2 z=0
\)