$A$ 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:

If the plane $\frac{x}{2}+\frac{y}{3}+\frac{z}{6}=1$ cuts the coordinate axes at points $A, B, C$ respectively,then the area of the triangle $ABC$ is

Let $\pi_1$ be the plane passing through the point $2\hat{i}-\hat{j}+\hat{k}$ and perpendicular to the vector $a\hat{i}+2\hat{j}-3\hat{k}$, and $\pi_2$ be the plane passing through the point $\hat{i}+2\hat{j}-\hat{k}$ and perpendicular to the vector $\hat{i}-2\hat{j}+\hat{k}$. If $\theta$ is the angle between the planes $\pi_1$ and $\pi_2$ and $\cos \theta = -\sqrt{\frac{3}{7}}$, then the integral value of $a$ is:

$A$ plane meets the coordinate axes at $A, B,$ and $C$ such that the centroid of triangle $ABC$ is $(1, 2, 3)$. Find the equation of the plane.

The equation of a plane containing the point $(1, -1, 2)$ and perpendicular to the planes $2x + 3y - 2z = 5$ and $x + 2y - 3z = 8$ is

If $-2, \frac{4}{3}, \frac{-4}{5}$ are the intercepts made by a plane on $X, Y, Z$-axes respectively,then the direction cosines of a normal to this plane are