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

The equation of a plane containing the lines $\overline{r}=(\hat{\imath}+2 \hat{\jmath}-4 \hat{k})+\lambda(2 \hat{\imath}+3 \hat{\jmath}+6 \hat{k})$ and $\overline{r}=(\hat{\imath}+3 \hat{\jmath}+4 \hat{k})+\mu(\hat{\imath}+\hat{\jmath}-\hat{k})$ is

The angle between the line $r = (2i - j + k) + \lambda (-i + j + k)$ and the plane $r \cdot (3i + 2j - k) = 4$ is:

The distance of the point $(1, 6, 2)$ from the point of intersection of the line $\frac{x-2}{3} = \frac{y+1}{4} = \frac{z-2}{12}$ and the plane $x-y+z=16$ is (in $\text{ units}$)

Show that the points $(\hat{i}-\hat{j}+3 \hat{k})$ and $3(\hat{i}+\hat{j}+\hat{k})$ are equidistant from the plane $\vec{r} \cdot(5 \hat{i}+2 \hat{j}-7 \hat{k})+9=0$ and lie on opposite sides of it.

For real numbers $\alpha$ and $\beta \neq 0$,if the point of intersection of the straight lines $\frac{x-\alpha}{1}=\frac{y-1}{2}=\frac{z-1}{3}$ and $\frac{x-4}{\beta}=\frac{y-6}{3}=\frac{z-7}{3}$ lies on the plane $x+2y-z=8$,then $\alpha-\beta$ is equal to: