Find the Cartesian equation of the plane $\vec{r} = (1 + \lambda - \mu)\hat{i} + (2 - \lambda)\hat{j} + (3 - 2\lambda + 2\mu)\hat{k}$.

The equation of a plane parallel to the plane $x - 2y + 2z - 5 = 0$ and at a unit distance from the origin is:

The equation of the plane which is parallel to the plane $x - 2y + 2z = 5$ and whose distance from the point $(1, 2, 3)$ is $1$ is:

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

Direction ratios of the normal to the plane passing through the points $A(0, 1, 1)$,$B(1, 1, 2)$,and $C(-1, 2, -2)$ are

If $(x, y, z)$ is an arbitrary point lying on a plane $P$ which passes through the points $(42, 0, 0)$,$(0, 42, 0)$,and $(0, 0, 42)$,then the value of the expression $3 + \frac{x-11}{(y-19)^{2}(z-12)^{2}} + \frac{y-19}{(x-11)^{2}(z-12)^{2}} + \frac{z-12}{(x-11)^{2}(y-19)^{2}} - \frac{x+y+z}{14(x-11)(y-19)(z-12)}$ is: