The line \(\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}\) lies in the plane \(x+3 y-\alpha z+\beta=0\), then the value of \(\alpha^2+\alpha \beta+\beta^2\) is

Line \(\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}\) lies in the plane \(x+3 y-\alpha z+\beta=0\)

The direction ratios of the line are \(3,-5,2\).
The direction ratios of the normal to the plane are \(1,3,-\alpha\).

The given line is perpendicular to the normal of plane.
\(\begin{aligned}
\therefore \quad & 3(1)+(-5)(3)+2(-\alpha)=0 \\
& \Rightarrow 3-15-2 \alpha=0 \\
& \Rightarrow-12-2 \alpha=0 \\
& \Rightarrow \alpha=-6
\end{aligned}\)
Also, point \((2,1,-2)\) lies on the plane
\(\begin{aligned}
& x+3 y-\alpha z+\beta=0 \\
& \Rightarrow 2+3-(-6)(-2)+\beta=0 \\
& \Rightarrow 2+3-12+\beta=0 \\
& \Rightarrow \beta=7 \\
\therefore \quad & \alpha^2+\alpha \beta+\beta^2=36-42+49=43
\end{aligned}\)