The equation of the plane containing the line \(\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}\) and the point \((0,7,-7)\) is

Let \(a, b, c\) be the direction cosines of the required plane.
It contains the line \(\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}\) and passes through the point \((0,7,-7)\)
\(\therefore \mathrm{a}(\mathrm{x}+1)+\mathrm{b}(\mathrm{y}-3)+\mathrm{c}(\mathrm{z}+2)=0 \quad \ldots(1) \)
\( \therefore \mathrm{a}(0+1)+\mathrm{b}(7-3)+\mathrm{c}(-7+2)=0 \Rightarrow \mathrm{a}+4 \mathrm{b}~-\) \(5 \mathrm{c}=0.\)
\(\text {Also }-3 \mathrm{a}+2 \mathrm{b}+\mathrm{c}=0
\)
From (2) and (3), we write
\(\frac{\mathrm{a}}{\left|\begin{array}{cc}
4 & -5 \\
2 & 1
\end{array}\right|}=\frac{-b}{\left|\begin{array}{cc}1 & -5 \\
-3 & 1
\end{array}\right|}=\frac{\mathrm{c}}{\left|\begin{array}{cc}1 & 4 \\
-3 & 2
\end{array}\right|}\)
\(\therefore \frac{\mathrm{a}}{14}=\frac{\mathrm{b}}{14}=\frac{\mathrm{c}}{14} \Rightarrow \mathrm{a}=\mathrm{b}=\mathrm{c}=1\)
Hence eq. (1) becomes
\(
x+1+y-3+z+2=0 \Rightarrow x+y+z=0
\)