Let the line $L$ pass through the point $(0,1,2)$,intersect the line $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and be parallel to the plane $2x+y-3z=4$. Then the distance of the point $P(1,-9,2)$ from the line $L$ is

If the three planes $x = 5$,$2x - 5ay + 3z - 2 = 0$,and $3bx + y - 3z = 0$ pass through a common line,then the value of $(a, b)$ is:

The point $P$ is the intersection of the straight line joining the points $Q(2, 3, 5)$ and $R(1, -1, 4)$ with the plane $5x - 4y - z = 1$. If $S$ is the foot of the perpendicular drawn from the point $T(2, 1, 4)$ to $QR$,then the length of the line segment $PS$ is

The square of the distance of the point of intersection of the line $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z+1}{6}$ and the plane $2x-y+z=6$ from the point $(-1,-1,2)$ is .... .

$A$ plane is making intercepts $2, 3, 4$ on $X, Y$ and $Z$-axes respectively. Another plane is passing through the point $(-1, 6, 2)$ and is perpendicular to the line joining the points $(1, 2, 3)$ and $(-2, 3, 4)$. Then the angle between the two planes is

The acute angle $\theta$ between the line $\vec{r} = (\hat{i} + 2\hat{j} + \hat{k}) + \lambda(\hat{i} + \hat{j} + \hat{k})$ and the plane $\vec{r} \cdot (2\hat{i} + p\hat{j} + \hat{k}) = 8$ is given by $\sin \theta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}$, where $\vec{b} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{n} = 2\hat{i} + p\hat{j} + \hat{k}$. If $\theta = \sin^{-1} \left( \frac{\sqrt{2}}{3} \right)$, then the value$(s)$ of $p$ is/are: