Question

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

Answer 1

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

to $3x-y-2z=7$

for $\perp a_1{a}_2+b_1{b}_2+c_1{c}_2=0$

$36-\lambda +16=0$

$\lambda =52$