Find the Cartesian equation of the following plane: $\vec{r} \cdot (2\hat{i} + 3\hat{j} - 4\hat{k}) = 1$

Let $\pi$ be the plane passing through the point $(3,-3,1)$ and perpendicular to the line joining the points $(3,4,-1)$ and $(2,-1,5)$. If the equation of the plane containing the points $(3,4,-1),(-1,2,5)$ and perpendicular to the plane $\pi$ is $ax+y+cz-d=0$,then $3(a+c)=$

The plane passing through $(2, 1, -3)$ and perpendicular to $3 \hat{i} - \hat{j} + 2 \hat{k}$ contains the points

The plane $\frac{x}{2}+\frac{y}{3}+\frac{z}{4}=1$ cuts the $X$-axis at $A$,$Y$-axis at $B$,and $Z$-axis at $C$. The area of $\triangle ABC$ is:

Find the angle between the planes $ax + by + d = 0$ $(a^2 + b^2 \neq 0)$ and $z = 0$.

Let $A=(-3,-2,7)$ and $B=(3,1,-2)$. Let a plane perpendicular to the line segment $AB$ divide $AB$ in the ratio $2:1$. Then the intercept made by the plane on the $y$-axis is