$A$ line with positive direction cosines passes through the point $P(2,-1,2)$ and makes equal angles with the coordinate axes. The line meets the plane $2x+y+z=9$ at point $Q$. The length of the line segment $PQ$ equals

The angle between the plane $x + y + z = 5$ and the line of intersection of the planes $3x + 4y + z - 1 = 0$ and $5x + 8y + 2z + 14 = 0$ is

If the lines $\frac{x - a + d}{\alpha - \delta} = \frac{y - a}{\alpha} = \frac{z - a - d}{\alpha + \delta}$ and $\frac{x - b + c}{\beta - \gamma} = \frac{y - b}{\beta} = \frac{z - b - c}{\beta + \gamma}$ are coplanar,then the equation of the plane containing them is .........

Let the equation of the plane passing through the line of intersection of the planes $x+2y+az=2$ and $x-y+z=3$ be $5x-11y+bz=6a-1$. For $c \in \mathbb{Z}$,if the distance of this plane from the point $(a, -c, c)$ is $\frac{2}{\sqrt{a}}$,then $\frac{a+b}{c}$ is equal to

In $\mathbb{R}^3$,let $L$ be a straight line passing through the origin. Suppose that all the points on $L$ are at a constant distance from the two planes $P_1: x+2y-z+1=0$ and $P_2: 2x-y+z-1=0$. Let $M$ be the locus of the feet of the perpendiculars drawn from the points on $L$ to the plane $P_1$. Which of the following points lie$(s)$ on $M$?
$(A) \left(0, -\frac{5}{6}, -\frac{2}{3}\right)$
$(B) \left(-\frac{1}{6}, -\frac{1}{3}, \frac{1}{6}\right)$
$(C) \left(-\frac{5}{6}, 0, \frac{1}{6}\right)$
$(D) \left(-\frac{1}{3}, 0, \frac{2}{3}\right)$

The distance of the point $P(1, 0, -3)$ from the plane $x - y - z = 9$ measured parallel to the line $\frac{x - 2}{2} = \frac{y + 2}{3} = \frac{z - 6}{-6}$ is (in $\text{ units}$)