What is the distance between the line $\frac{x - 1}{3} = \frac{y + 2}{-2} = \frac{z - 1}{2}$ and the plane $2x + 2y - z = 6$?

The line of shortest distance between the lines $\frac{x-2}{0}=\frac{y-1}{1}=\frac{z}{1}$ and $\frac{x-3}{2}=\frac{y-5}{2}=\frac{z-1}{1}$ makes an angle of $\cos ^{-1}\left(\sqrt{\frac{2}{27}}\right)$ with the plane $P: ax-y-z=0$,$(a>0)$. If the image of the point $(1,1,-5)$ in the plane $P$ is $(\alpha, \beta, \gamma)$,then $\alpha+\beta-\gamma$ is equal to $........$

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

If the three planes $x = 5, 2x - 5ay + 3z - 2 = 0$ and $3bx + y - 3z = 0$ contain a common line,then $(a, b)$ is equal to

Let $L$ be the line of intersection of the planes $2x + 3y + z = 1$ and $x + 3y + 2z = 2$. If $L$ makes an angle $\alpha$ with the positive direction of the $x$-axis,then $\cos \alpha$ is:

Let the lines $L_{1}: \overrightarrow{r} = \lambda(\hat{i} + 2\hat{j} + 3\hat{k}), \lambda \in R$ and $L_{2}: \overrightarrow{r} = (\hat{i} + 3\hat{j} + \hat{k}) + \mu(\hat{i} + \hat{j} + 5\hat{k}), \mu \in R$ intersect at the point $S$. If a plane $ax + by - z + d = 0$ passes through $S$ and is parallel to both the lines $L_{1}$ and $L_{2}$,then the value of $a + b + d$ is equal to: