The equation of the perpendicular line from the point $(2,-3,1)$ to the line $\dfrac{x + 1}{2} = \dfrac{y - 3}{3} = \dfrac{z + 2}{-1}$ is

The angle between the line $x - 1 = 2 - y = \dfrac{2z - 6}{4}$ and the plane $\vec{r} \cdot (2\hat{i} + \hat{j} + \hat{k}) = 10$ is

The sum of the coordinates of one of the points on the line $\dfrac{x - 2}{1} = \dfrac{y + 3}{-2} = \dfrac{z + 5}{2}$ which is at a distance of 3 units from the point $(2, -3, -5)$ is ...

If A and B are the feet of the perpendiculars drawn from $(1, 2, 3)$ to planes $YZ$ and $ZX$, then the equation of the plane passing through the points A, B and the origin is