The direction ratios of the line of intersection of the planes $x-y+z-5=0$ and $x-3y-6=0$ are:

Let the plane $2x + 3y + z + 20 = 0$ be rotated through a right angle about its line of intersection with the plane $x - 3y + 5z = 8$. If the mirror image of the point $(2, -1/2, 2)$ in the rotated plane is $B(a, b, c)$,then:

Three lines are given by $\overrightarrow{r} = \lambda \hat{i}, \lambda \in R$,$\overrightarrow{r} = \mu(\hat{i} + \hat{j}), \mu \in R$ and $\overrightarrow{r} = v(\hat{i} + \hat{j} + \hat{k}), v \in R$. Let the lines cut the plane $x + y + z = 1$ at the points $A, B$ and $C$ respectively. If the area of the triangle $ABC$ is $\Delta$,then the value of $(6 \Delta)^2$ equals.

Find the equation of the plane passing through the intersection of the planes $x + 2y + 3z = 4$ and $2x + y - z = -5$,and perpendicular to the plane $5x + 3y + 6z + 8 = 0$.

If the line $\vec{r} = (\hat{i} - 2\hat{j} - \hat{k}) + \lambda (2\hat{i} + \hat{j} + 2\hat{k})$ is parallel to the plane $\vec{r} \cdot (3\hat{i} - 2\hat{j} - m\hat{k}) = 14$,then the value of $m$ is:

If the line,$\frac{x-3}{2}=\frac{y+2}{1}=\frac{z+4}{3}$ lies in the plane,$\ell x+m y-z=9$,then $\ell^2+m^2$ is equal to