• # question_answer   A plane P is perpendicular to the vector $\vec{A}=2\hat{i}+3\hat{j}+6\hat{k}$ and contains the terminal point of the vector $\vec{B}=\hat{i}+5\hat{j}+3\hat{k}$. The distance from the origin to the plane P, is A)  1                     B)  2 C)  4                                 D)  5

Plane P passes through (1, 5, 3) and normal to $\vec{A}=(2,3,6)$ Equation is $2(x-1)+3(y-5)\,+6(z-3)\,=0$$\Rightarrow \,2x+3y+6z=35$ $\therefore$ Perpendicular from (0, 0, 0) to the plane $=\left| \frac{-35}{7} \right|=5$