Let the line $\frac{x - 2}{3} = \frac{y - 1}{-5} = \frac{z + 2}{2}$ lie in the plane $x + 3y - \alpha z + \beta = 0$. Then $(\alpha, \beta)$ equals.

Find the coordinates of the point where the line passing through $(3, -4, -5)$ and $(2, -3, 1)$ intersects the plane $2x + y + z = 7$.

Let two planes be given by $P_1 : 2x - y + z = 2$ and $P_2 : x + 2y - z = 3$. Based on the given information,find the equation of the plane passing through the intersection of $P_1$ and $P_2$ and the point $(3, 2, 1)$.

The line of intersection of the planes $r \cdot (i - 3j + k) = 1$ and $r \cdot (2i + 5j - 3k) = 2$ is parallel to the vector

The equation of the perpendicular from the point $(\alpha, \beta, \gamma)$ to the plane $ax + by + cz + d = 0$ is

The line passing through the point $2\bar{a}+\bar{b}$ and parallel to the vector $\bar{b}-\bar{c}$ and the plane passing through the point $\bar{a}$ and parallel to the vectors $\bar{b}+\bar{c}$ and $\bar{a}+2\bar{b}-\bar{c}$ intersect at $P$. The position vector of $P$ is