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.

The foot of perpendicular from the origin to the given plane is

$\left(\frac{2}{15}, \frac{2}{75},-\frac{11}{75}\right)$

Let foot of perpendicular = (p, q, r)

so   from (0, 0, 0)

so  $\frac{p-0}{10}=\frac{q-0}{2}=\frac{r-0}{-11}=\frac{-(-3)}{\left(10^2+2^2+(-11)^2\right)}$

$\Rightarrow \frac{p}{10}=\frac{q}{2}=\frac{r}{-11}=\frac{3}{225} \Rightarrow (p, q, r)$

$=\left(\frac{30}{225}, \frac{6}{225}, \frac{-33}{225}\right)$

$=\left(\frac{2}{15}, \frac{2}{75}, \frac{-11}{75}\right)$

Option: B