The point of intersection of the line $\frac{x}{1} = \frac{y - 1}{2} = \frac{z + 2}{3}$ and the plane $2x + 3y + z = 0$ is

The value of $m$,such that the line $\frac{x-4}{1}=\frac{y-2}{1}=\frac{2z-m}{3}$ lies in the plane $2x-5y+2z=7$,is

Let $P(3, 2, 6)$ be a point in space and $Q$ be a point on the line $\vec{r} = (\hat{i} - \hat{j} + 2\hat{k}) + \mu(-3\hat{i} + \hat{j} + 5\hat{k})$. Then the value of $\mu$ for which the vector $\vec{PQ}$ is parallel to the plane $x - 4y + 3z = 1$ is

Find the equation of the line passing through the point $(3, 0, 1)$ and parallel to the planes $x+2y=0$ and $3y-z=0$.

If the points $(1, 1, \lambda)$ and $(-3, 0, 1)$ are equidistant from the plane $3x + 4y - 12z + 13 = 0$,then the integer value of $\lambda$ is:

Let the plane containing the line of intersection of the planes $P_1: x+(\lambda+4)y+z=1$ and $P_2: 2x+y+z=2$ pass through the points $(0,1,0)$ and $(1,0,1)$. Then the distance of the point $(2\lambda, \lambda, -\lambda)$ from the plane $P_2$ is (in $\sqrt{6}$)