The equation of the plane passing through $(1, 2, 3)$ and parallel to the plane $2x + 3y - 4z = 0$ is

In three-dimensional space,the equation $3y + 4z = 0$ represents:

$\pi_1$ is a plane passing through the point $(1, 2, 3)$ and perpendicular to the planes $x+2y+3z-6=0$ and $x+2y+2z-5=0$. If $(-1, 2, -3)$ is the foot of the perpendicular drawn from the point $(1, 3, 2)$ onto a plane $\pi_2$, then the angle between the planes $\pi_1$ and $\pi_2$ is

The direction ratios of the normal to the plane passing through the points $(1, 0, 0)$ and $(0, 1, 0)$ and making an angle of $\pi/4$ with the plane $x + y = 3$ are proportional to:

If the cartesian equation of the plane passing through the point $\hat{i}+2 \hat{j}+\hat{k}$ and parallel to the vectors $2 \hat{i}+3 \hat{j}+\hat{k}$ and $-\hat{i}+2 \hat{j}-3 \hat{k}$ is $a x+b y+c z=1$, then $18(a+b+c)$ is equal to

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: