The distance of the point $(2, 3, -5)$ from the plane $x + 2y - 2z = 9$ is

Let the equation of the plane $P$ containing the line $x+10=\frac{8-y}{2}=z$ be $ax+by+3z=2(a+b)$ and the distance of the plane $P$ from the point $(1,27,7)$ be $c$. Then $a^2+b^2+c^2$ is equal to $.............$.

Let $P$ be the plane passing through the intersection of the planes $\overrightarrow{r} \cdot (\hat{i} + 3\hat{j} - \hat{k}) = 5$ and $\overrightarrow{r} \cdot (2\hat{i} - \hat{j} + \hat{k}) = 3$,and the point $(2, 1, -2)$. Let the position vectors of the points $X$ and $Y$ be $\hat{i} - 2\hat{j} + 4\hat{k}$ and $5\hat{i} - \hat{j} + 2\hat{k}$ respectively. Then the points:

$A$ 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

The angle between the line $\frac{x+1}{2}=\frac{y-2}{1}=\frac{z-3}{-2}$ and the plane $x-2y-\lambda z=3$ is $\cos^{-1}\left(\frac{2\sqrt{2}}{3}\right)$. Then the value of $\lambda$ is:

The equation of the plane containing the lines $\frac{x - 1}{2} = \frac{y + 1}{\lambda} = \frac{z}{2}$ and $\frac{x + 1}{5} = \frac{y + 1}{2} = \frac{z}{\lambda}$ is