Suppose the line $\frac{x-2}{\alpha}=\frac{y-2}{-5}=\frac{z+2}{2}$ lies on the plane $x+3y-2z+\beta=0$. Then $(\alpha+\beta)$ is equal to ... .

The foot of the perpendicular drawn from the point $(4,2,3)$ to the line joining the points $(1,-2,3)$ and $(1,1,0)$ lies on the plane

If the distance between the plane $ax - 2y + z = k$ and the plane containing the lines $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$ and $\frac{x-2}{3} = \frac{y-3}{4} = \frac{z-4}{5}$ is $\sqrt{6}$, then $|k|$ is

Let $\overrightarrow{a} = \hat{i} + 2\hat{j} + \hat{k}$ and $\overrightarrow{b} = 2\hat{i} + 7\hat{j} + 3\hat{k}$. Let $L_1: \overrightarrow{r} = (-\hat{i} + 2\hat{j} + \hat{k}) + \lambda \overrightarrow{a}, \lambda \in R$ and $L_2: \overrightarrow{r} = (\hat{j} + \hat{k}) + \mu \overrightarrow{b}, \mu \in R$ be two lines. If the line $L_3$ passes through the point of intersection of $L_1$ and $L_2$,and is parallel to $\overrightarrow{a} + \overrightarrow{b}$,then $L_3$ passes through the point:

The equation of the plane containing the lines $\frac{x - 1}{2} = \frac{y + 1}{\lambda} = \frac{z}{2}$ and $\frac{x + 1}{5} = \frac{y + 1}{2} = \frac{z}{\lambda}$ is

The distance of the point $A(3, -4, 5)$ from the plane $2x + 5y - 6z = 16$ measured along the line $\frac{x}{2} = \frac{y}{1} = \frac{z}{-2}$ is