The equation of the plane containing the lines $r = a_1 + \lambda a_2$ and $r = a_2 + \lambda a_1$ is

The plane passing through the point $(4, -1, 2)$ and parallel to the lines $\frac{x + 2}{3} = \frac{y - 2}{-1} = \frac{z + 1}{2}$ and $\frac{x - 2}{1} = \frac{y - 3}{2} = \frac{z - 4}{3}$ also passes through the point

The foot of the perpendicular drawn from the point $(-2,-1,3)$ to a plane $\pi$ is $(1,0,-2)$. If $a, b, c$ are the intercepts made by the plane $\pi$ on $X, Y, Z$-axes respectively, then $3a+b+5c=$

The distance of a point $(2, 3, -5)$ from the plane $\vec{r} \cdot (4 \hat{i} - 3 \hat{j} + 2 \hat{k}) = 4$ is

Let the plane passing through the point $(-1, 0, -2)$ and perpendicular to each of the planes $2x + y - z = 2$ and $x - y - z = 3$ be $ax + by + cz + 8 = 0$. Then the value of $a + b + c$ is equal to:

$A$ plane meets the coordinate axes at $A, B, C$ such that the centroid of the $\Delta ABC$ is the point $(\alpha, \beta, \gamma)$. Show that the equation of the plane is $\frac{x}{\alpha} + \frac{y}{\beta} + \frac{z}{\gamma} = 3$.