Find the equation of the plane passing through the origin and parallel to the plane $3x - 4y + 5z - 6 = 0$.

If the point $(2, \alpha, \beta)$ lies on the plane which passes through the points $(3, 4, 2)$ and $(7, 0, 6)$ and is perpendicular to the plane $2x - 5y = 15$,then $2\alpha - 3\beta$ is equal to:

Find the equation of the plane passing through the point $(1, 2, 3)$ and parallel to the plane $2x + 3y - 4z = 0$.

The equation of the plane passing through a point $A(2, -1, 3)$ and parallel to the vectors $\vec{a} = (3, 0, -1)$ and $\vec{b} = (-3, 2, 2)$ is:

The plane passing through $(2, 1, -3)$ and perpendicular to $3 \hat{i} - \hat{j} + 2 \hat{k}$ contains the points

If the distance between the plane,$23x - 10y - 2z + 48 = 0$ and the plane containing the lines $\frac{x+1}{2} = \frac{y-3}{4} = \frac{z+1}{3}$ and $\frac{x+3}{2} = \frac{y+2}{6} = \frac{z-1}{\lambda}$ $(\lambda \in R)$ is equal to $\frac{k}{\sqrt{633}}$,then $k$ is equal to