The plane containing the point $(3,2,0)$ and the line $\frac{x-3}{1}=\frac{y-6}{5}=\frac{z-4}{4}$ is

Let the lines $l_1: \frac{x+5}{3}=\frac{y+4}{1}=\frac{z-\alpha}{-2}$ and $l_2: 3x+2y+z-2=0=x-3y+2z-13$ be coplanar. If the point $P(a, b, c)$ on $l_1$ is nearest to the point $Q(-4, -3, 2)$,then $|a|+|b|+|c|$ is equal to

The direction cosines of the line of intersection of the planes $x+2y+z-4=0$ and $2x-y+z-3=0$ are

The equation of the plane passing through the line of intersection of the planes $x + y + z = 5$ and $2x + 3y + 4z + 5 = 0$ and perpendicular to the plane $x + y + z = 5$ is

Let $L$ be a line obtained from the intersection of two planes $x+2y+z=6$ and $y+2z=4$. If point $P(\alpha, \beta, \gamma)$ is the foot of the perpendicular from $(3,2,1)$ on $L$,then the value of $21(\alpha+\beta+\gamma)$ equals ...... .

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