By using equations of the line $\frac{x+1}{2}=\frac{y}{3}=\frac{z-3}{6}$ and the plane $10 x+2 y-11 z-3=0$, answer the following questions.

Co-ordinates of point of intersection of line and plane

$\left(\frac{-33}{10}, \frac{-69}{20}, \frac{-39}{10}\right)$

let $\frac{x+1}{2}=\frac{y}{3}=\frac{z-3}{6}=\lambda \Rightarrow  x=2 \lambda-1 , y=3 \lambda, z=6 \lambda+3$

Putting in eqⁿ of plane

$10(2 \lambda-1)+z(3 \lambda)-11(6 \lambda+3)-3=0$

$\Rightarrow -40 \lambda-46=0 \Rightarrow \lambda=\frac{-23}{20}$

So  $x=\frac{-46}{20} -1~~~ y=-\frac{69}{20} ~~~z=\frac{-69}{10}+3$

$\Rightarrow x=\frac{-33}{10} ~~~y=\frac{-69}{20} ~~~z=-\frac{39}{10}$

Option: B