If the equation of the plane passing through the point $(2, -1, 3)$ and perpendicular to the planes $3x - 2y + z = 9$ and $x + y + z = 9$ is $x + by + cz + d = 0$, then $d =$

If the line $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 2}{4}$ meets the plane $x + 2y + 3z = 15$ at a point $P$,then the distance of $P$ from the origin is

Three lines $L_1: \overrightarrow{r} = \lambda \hat{i}, \lambda \in R$,$L_2: \overrightarrow{r} = \hat{k} + \mu \hat{j}, \mu \in R$,and $L_3: \overrightarrow{r} = \hat{i} + \hat{j} + v\hat{k}, v \in R$ are given. For which point$(s)$ $Q$ on $L_2$ can we find a point $P$ on $L_1$ and a point $R$ on $L_3$ such that $P, Q,$ and $R$ are collinear?

The line of intersection of the planes $\vec r \cdot (3\hat i - \hat j + \hat k) = 1$ and $\vec r \cdot (\hat i + 4\hat j - 2\hat k) = 2$ is:

$A$ line with positive 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 $PQ$ equals

If the equation of the plane passing through the mirror image of a point $(2,3,1)$ with respect to the line $\frac{x+1}{2}=\frac{y-3}{1}=\frac{z+2}{-1}$ and containing the line $\frac{x-2}{3}=\frac{1-y}{2}=\frac{z+1}{1}$ is $\alpha x+\beta y+\gamma z=24$,then $\alpha+\beta+\gamma$ is equal to ..... .