The equation of a plane containing the point \((1,-1,2)\) and perpendicular to the
planes \(2 x+3 y-2 z=5\) and \(x+2 y-3 z=8\) is

(A)
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 \(\overline{\mathrm{r}} \cdot(5 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}-\hat{\mathrm{k}})=7\)