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

The length of the projection of the line segment joining the points $(5, -1, 4)$ and $(4, -1, 3)$ on the plane $x + y + z = 7$ is:

Let the foot of the perpendicular from the point $P (3, -2, -9)$ on the plane passing through the points $A (-1, -2, -3)$,$B (9, 3, 4)$,and $C (9, -2, 1)$ be $Q(\alpha, \beta, \gamma)$. Then the distance of $Q$ from the origin is:

The line $\frac{x + 3}{3} = \frac{y - 2}{-2} = \frac{z + 1}{1}$ and the plane $4x + 5y + 3z - 5 = 0$ intersect at a point

In $R^3$,consider the planes $P_1: y=0$ and $P_2: x+z=1$. Let $P_3$ be a plane,different from $P_1$ and $P_2$,which passes through the intersection of $P_1$ and $P_2$. If the distance of the point $(0,1,0)$ from $P_3$ is $1$ and the distance of a point $(\alpha, \beta, \gamma)$ from $P_3$ is $2$,then which of the following relations is (are) true?
$(A)$ $2\alpha+\beta+2\gamma+2=0$
$(B)$ $2\alpha-\beta+2\gamma+4=0$
$(C)$ $2\alpha+\beta-2\gamma-10=0$
$(D)$ $2\alpha-\beta+2\gamma-8=0$

If the distance of the point $P(1, -2, 1)$ from the plane $x + 2y - 2z = \alpha$, where $\alpha > 0$, is $5$, then the foot of the perpendicular from $P$ to the plane is