Find the ratio in which the plane $2x + 3y + 5z = 1$ divides the line segment joining the points $(1, 0, -3)$ and $(1, -5, 7)$.

The equation of the plane passing through the intersection of the lines $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-5}{-3}$ and $\frac{x+5}{3}=\frac{y-4}{-1}=\frac{z+3}{4}$ and parallel to the $xy$-plane is

In what ratio does the plane $ax + by + cz + d = 0$ divide the line segment joining the points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$?

The distance between the line $r = 2i - 2j + 3k + \lambda (i - j + 4k)$ and the plane $r \cdot (i + 5j + k) = 5$ is

If $P$,$Q$,and $R$ are the feet of the perpendiculars drawn from the point $A(1, 1, 1)$ to the planes $P_1: x + 2y + 2z = 2$,$P_2: 2x - 2y + z = -8$,and to the line of intersection of $P_1$ and $P_2$ respectively,then the area of $\Delta PQR$ is:

Let $P_{1}: \vec{r} \cdot(2 \hat{i} + \hat{j} - 3 \hat{k}) = 4$ be a plane. Let $P_{2}$ be another plane which passes through the points $(2, -3, 2)$,$(2, -2, -3)$,and $(1, -4, 2)$. If the direction ratios of the line of intersection of $P_{1}$ and $P_{2}$ are $16, \alpha, \beta$,then the value of $\alpha + \beta$ is equal to