The vector equation of the plane containing the lines $r = (i + j) + \lambda (i + 2j - k)$ and $r = (i + j) + \mu (-i + j - 2k)$ 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

The plane $3x + 4y + 6z + 7 = 0$ is rotated about the line $r = (\hat{i} + 2\hat{j} - 3\hat{k}) + t(2\hat{i} - 3\hat{j} + \hat{k})$ until the plane passes through the origin. The equation of the plane in the new position is

Let the lines $L_1: \vec{r}=\hat{i}+2\hat{j}+3\hat{k}+\lambda(2\hat{i}+3\hat{j}+4\hat{k})$, $\lambda \in R$ and $L_2: \vec{r}=(4\hat{i}+\hat{j})+\mu(5\hat{i}+2\hat{j}+\hat{k})$, $\mu \in R$, intersect at the point $R$. Let $P$ and $Q$ be the points lying on lines $L_1$ and $L_2$, respectively, such that $|\overrightarrow{PR}|=\sqrt{29}$ and $|\overrightarrow{PQ}|=\sqrt{\frac{47}{3}}$. If the point $P$ lies in the first octant, then $27(QR)^2$ is equal to

The length of the projection of the line segment joining the points $(5,-1,4)$ and $(4,-1,3)$ on the plane $x+y+z=7$ is

The plane containing the point $(3,2,0)$ and the line $\frac{x-3}{1}=\frac{y-6}{5}=\frac{z-4}{4}$ is