The equation of a plane passing through $(1,-1,2)$ is $a(x-1)+b(y+1)+c(z-2)=0$
If is perpendicular to the planes $2 x+3 y-2 z=5$ and $x+2 y-3 z=8$.
$\therefore 2 \mathrm{a}+3 \mathrm{~b}-2 \mathrm{c}=0$ and $\mathrm{a}+2 \mathrm{~b}-3 \mathrm{c}=0$
Solving the above equation, we get,
$\begin{array}{l}
\frac{a}{\left|\begin{array}{cc}
3 & -2 \\
2 & -3
\end{array}\right|}=\frac{-b}{\left|\begin{array}{ll}
2 & -2 \\
1 & -3
\end{array}\right|}=\frac{c}{\left|\begin{array}{ll}
2 & 3 \\
1 & 2
\end{array}\right|} \\
\frac{a}{-5}=\frac{b}{4}=\frac{c}{1}
\end{array}$
Substituting $\mathrm{a}=-5, \mathrm{~b}=4$ and $\mathrm{c}=1$, we get, $-5 x+4 y+z=-7 \Rightarrow 5 x-4 y-z=7$
$\therefore$ Required equation can be written as $\overline{\mathrm{r}} \cdot(5 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}-\hat{\mathrm{k}})=7$