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 value…
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 value of $\alpha \beta$ is
42
1
-42
-2
Solution
line $\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$ lies in the plane $x+3 y-\alpha z+\beta=0$
$\therefore(3)(1)+(-5)(3)+2(-\alpha)=0 \Rightarrow \alpha=-6$
Thus equation of plane is $x+3 y+6 z+\beta=0$ and point $(2,1,-2)$ lies in it.
$\begin{aligned}
& \therefore 2+3(1)+6(-2)+\beta=0 \Rightarrow \beta=7 \\
& \therefore \alpha \beta=(-6)(7)=-42
\end{aligned}$