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.

Distance of plane 10x + 2y – 11z – 3 = 0, from the origin is:

1 $\frac{1}{10}$ $\frac{1}{15}$ $\frac{1}{5}$

$\frac{1}{5}$

O(0, 0, 0) Distance = $\frac{|10 \times 0+2 \times 0-11 \times 0-3|}{\sqrt{10^2+2^2+(-11)^2}}=\frac{3}{15}=\frac{1}{5}$ Option: D