Find the value of $\lambda$ such that the line $\frac{x-2}{12}=\frac{y-1}{\lambda }=\frac{z-3}{-8}$ is perpendicular to the plane $3x-y-2z=7$.

Given, equation of line is $\frac{x-2}{12}=\frac{y-1}{\lambda }=\frac{z-3}{-8}$. ...........(1)

and equation of plane is $3x-y-2z=7$ .........(2)

On comparing equation 1 with $\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z_{-}{z}_1}{c_1}$, we get,

$a_1=12,b_1=\lambda ,c_1=-8$

On comparing equation 2 with $a_{2}x+b_{2}y+c_{2}z=d_2$, we get,

$a_2=3,b_2=-1,c_2=-2,d_2=7$

Since, the given line is perpendicular to given plane.

So,

$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

$\frac{12}{3}=\frac{\lambda }{-1}=\frac{-8}{-2}$

Hence, $\lambda =-4$.