The angle between a normal to the plane $2x - y + 2z - 1 = 0$ and the $X$-axis is

$A$ vector $\overrightarrow{V}$ in the first octant is inclined to the $x$-axis at $60^{\circ}$,to the $y$-axis at $45^{\circ}$ and to the $z$-axis at an acute angle. If a plane passing through the points $(\sqrt{2}, -1, 1)$ and $(a, b, c)$ is normal to $\overrightarrow{V}$,then:

$A$ plane passes through the point $(1, -2, 1)$ and is perpendicular to the two planes $2x - 2y + z = 0$ and $x - y + 2z = 4$. Find the distance of the plane from the point $(1, 2, 2)$.

The equation of the plane in normal form which passes through the points $(-2,1,3), (1,1,1)$ and $(2,3,4)$ is

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:

If the position vectors of the points $A$ and $B$ are $3 \hat{i}+\hat{j}+2 \hat{k}$ and $\hat{i}-2 \hat{j}-4 \hat{k}$ respectively,then the equation of the plane passing through $B$ and perpendicular to $AB$ is