$A$ plane $ax+by+cz+1=0$ is perpendicular to the two planes $2x-2y+z=0$ and $x-y+2z=4$ and passes through the point $(1, -2, 1)$. Then $a+b-c=$

The vector equation of the plane passing through the intersection of the planes $\overrightarrow{r} \cdot (\hat{i} + \hat{j} + \hat{k}) = 1$ and $\overrightarrow{r} \cdot (\hat{i} - 2\hat{j}) = -2$,and the point $(1, 0, 2)$ is:

The acute angle between the line $\bar{r}=(\hat{i}+2\hat{j}+\hat{k})+\lambda(\hat{i}+\hat{j}+\hat{k})$ and the plane $\bar{r} \cdot(2\hat{i}-\hat{j}+\hat{k})=5$ is

For $a, b \in \mathbb{Z}$ and $|a - b| \leq 10$,let the angle between the plane $P: ax + y - z = b$ and the line $l: x - 1 = \frac{-y}{1} = z + 1$ be $\cos^{-1}\left(\frac{1}{3}\right)$. If the distance of the point $(6, -6, 4)$ from the plane $P$ is $3\sqrt{6}$,then $a^4 + b^2$ is equal to

$A$ line $L$ passes through the points $\hat{i}+2 \hat{j}+\hat{k}$ and $-2 \hat{i}+3 \hat{k}$. $A$ plane $P$ passes through the origin and the points $4 \hat{k}, 2 \hat{i}+\hat{j}$. The point where the line $L$ meets the plane $P$ is

$A$ plane which passes through the point $(3, 2, 0)$ and the line $\frac{x - 3}{1} = \frac{y - 6}{5} = \frac{z - 4}{4}$ is