Find the point of intersection of the line $\frac{x}{1} = \frac{y}{2} = \frac{z}{2}$ and the plane $2x + y + z = 6$.

The distance of the point $P(3,4,4)$ from the point of intersection of the line joining the points $Q(3,-4,-5)$ and $R(2,-3,1)$ with the plane $2x+y+z=7$ is: (in $units$)

The angle between the line $\frac{x - 1}{-2} = \frac{y - 2}{1} = \frac{z + 1}{2}$ and the plane $3x + 2y + 6z = 1$ is:

Let $d$ be the distance between the foot of perpendiculars of the points $P(1, 2, -1)$ and $Q(2, -1, 3)$ on the plane $-x + y + z = 1$. Then $d^{2}$ is equal to

$A$ plane $\pi$ is passing through the points $A(1, -2, 3)$ and $B(6, 4, 5)$. If the plane $\pi$ is perpendicular to the plane $3x - y + z = 2$,then the perpendicular distance from $(0, 0, 0)$ to the plane $\pi$ is

If the foot of the perpendicular from the point $A(-1, 4, 3)$ on the plane $P: 2x + my + nz = 4$ is $B\left(-2, \frac{7}{2}, \frac{3}{2}\right)$,then the distance of the point $A$ from the plane $P$,measured parallel to a line with direction ratios $3, -1, -4$,is equal to.