If the line $\frac{x - 1}{2} = \frac{y + \alpha}{\alpha} = \frac{z - \beta}{2}$ lies in the plane $2x + y + z = 5$,then $\alpha + \beta$ is

If the line $\frac{2x - 8}{\sin \beta} = \frac{y - \sin \alpha}{1} = \frac{z - 1}{\cos \alpha}$,where $\beta \in R$ and $\sin \beta \neq 1$,lies in the plane $2x - (\sin \beta)y + (\cos \beta)z = k$ for all $\alpha \in R$,then:

The plane $2x - 3y + 6z - 11 = 0$ makes an angle $\sin^{-1}(\alpha)$ with the $X$-axis. The value of $\alpha$ is equal to:

Let $PQR$ be a triangle with $R(-1, 4, 2)$. Suppose $M(2, 1, 2)$ is the midpoint of $PQ$. The distance of the centroid of $\triangle PQR$ from the point of intersection of the lines $\frac{x-2}{0} = \frac{y}{2} = \frac{z+3}{-1}$ and $\frac{x-1}{1} = \frac{y+3}{-3} = \frac{z+1}{1}$ is

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

The point $\bar{i}-2 \bar{j}$ lies on a line parallel to the vector $2 \bar{i}+\bar{k}$. The point $\bar{i}+2 \bar{j}$ lies on a plane parallel to the vectors $2 \bar{j}-\bar{k}$ and $\bar{i}+2 \bar{k}$. Find the point of intersection of the line and the plane.