The value of $k$,such that the line $\frac{x-4}{1}=\frac{y-2}{1}=\frac{z-k}{2}$ lies on the plane $2x-4y+z=7$,is

Find the coordinates of the foot of the perpendicular drawn from the origin to the plane $5y + 8 = 0$.

The equation of the plane $\pi$ through the line of intersection of the planes $\pi_1 \equiv x+3y-6=0$ and $\pi_2 \equiv 3x-y+4z=0$ is $\pi_1+\lambda \pi_2=0$. If the plane $\pi$ is at unit distance from the origin,then an equation of the plane $\pi$ is

The distance of the point $(-1, 9, -16)$ from the plane $2x + 3y - z = 5$ measured parallel to the line $\frac{x+4}{3} = \frac{2-y}{4} = \frac{z-3}{12}$ is $......$

The coordinates of the foot of the perpendicular from the origin $(0, 0, 0)$ to the plane $2x - 3y - 6z = 49$ are

The distance of the point $(3, 8, 2)$ from the line $\frac{x - 1}{2} = \frac{y - 3}{4} = \frac{z - 2}{3}$ measured parallel to the plane $3x + 2y - 2z = 0$ is