The distance between the planes $x + 2y + 3z + 7 = 0$ and $2x + 4y + 6z + 7 = 0$ is

Let three planes be:
$P_1 : x - y + z = 1$
$P_2 : x + y - z = -1$
$P_3 : x - 3y + 3z = 2$
Let $L_1, L_2, L_3$ be the lines of intersection of planes $(P_2, P_3)$,$(P_3, P_1)$,and $(P_1, P_2)$ respectively.
Statement-$1$: At least two of the lines $L_1, L_2, L_3$ are not parallel.
Statement-$2$: The three planes do not have a common point.

If the plane $2x - y + 2z + 3 = 0$ has distances of $\frac{1}{3}$ and $\frac{2}{3}$ units from the planes $4x - 2y + 4z + \lambda = 0$ and $2x - y + 2z + \mu = 0$ respectively,then the maximum value of $\lambda + \mu$ is equal to:

The intercepts of the plane $5x - 3y + 6z = 60$ on the coordinate axes are

For scalars $\lambda, \mu$,if the vector equation of a plane is $r=(2+3 \lambda-\mu) \hat{i}+(1-2 \lambda+3 \mu) \hat{j}+(-2+2 \lambda+\mu) \hat{k}$,then its Cartesian equation is

The equation of the plane passing through the point $(2, -1, -3)$ and parallel to the lines $\frac{x - 1}{3} = \frac{y + 2}{2} = \frac{z}{-4}$ and $\frac{x}{2} = \frac{y - 1}{-3} = \frac{z - 2}{2}$ is