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

If the plane passing through the points $\hat{i}+\hat{j}+\hat{k}$, $2\hat{i}-\hat{k}$ and the origin meets the line passing through the points $\hat{i}+3\hat{j}-2\hat{k}$ and $\hat{i}-\hat{j}+3\hat{k}$ at the point $A$, then $A=$

The distance of the point $P(1, 2, 1)$ from the plane $2x + y - z = 10$ measured along the line $\frac{x - 5}{1} = \frac{2y - 3}{2} = \frac{z - \frac{5}{2}}{1}$ is

Let $S$ be the reflection of a point $Q$ with respect to the plane given by $\vec{r} = -(t+p) \hat{i} + \hat{j} + (1+p) \hat{k}$,where $t, p$ are real parameters and $\hat{i}, \hat{j}, \hat{k}$ are the unit vectors along the three positive coordinate axes. If the position vectors of $Q$ and $S$ are $10 \hat{i} + 15 \hat{j} + 20 \hat{k}$ and $\alpha \hat{i} + \beta \hat{j} + \gamma \hat{k}$ respectively,then which of the following is/are $TRUE$?
$(A)$ $3(\alpha+\beta) = -101$
$(B)$ $3(\beta+\gamma) = -71$
$(C)$ $3(\gamma+\alpha) = -86$
$(D)$ $3(\alpha+\beta+\gamma) = -121$

Let two vertices of triangle $ABC$ be $(2,4,6)$ and $(0,-2,-5)$,and its centroid be $(2,1,-1)$. If the image of the third vertex in the plane $x+2y+4z=11$ is $(\alpha, \beta, \gamma)$,then $\alpha \beta+\beta \gamma+\gamma \alpha$ is equal to

The line $\frac{x - 2}{3} = \frac{y + 1}{2} = \frac{z - 1}{-1}$ intersects the curve $xy = c^2, z = 0$ if $c$ is equal to