The position vector of the point where the line $r = i - j + k + t(i + j - k)$ meets the plane $r \cdot (i + j + k) = 5$ is

Let $A$ be a point on the line $\vec{r} = (1 - 3\mu)\hat{i} + (\mu - 1)\hat{j} + (2 + 5\mu)\hat{k}$ and $B(3, 2, 6)$ be a point in space. Then the value of $\mu$ for which the vector $\overrightarrow{AB}$ is parallel to the plane $x - 4y + 3z = 1$ is

Find the distance of the point whose position vector is $(2 \hat{i}+\hat{j}-\hat{k})$ from the plane $\vec{r} \cdot(\hat{i}-2 \hat{j}+4 \hat{k})=9$.

If the line $\frac{x+1}{1}=\frac{y-k}{11}=\frac{z-4}{-5}$ lies in the plane $2x+py+7z-41=0$ which is perpendicular to the plane $x+4y-2z+13=0$,then $k=$

The Cartesian equation of a line passing through $(1, 2, 3)$ and parallel to the planes $x - y + 2z = 5$ and $3x + y + z = 6$ is:

The distance of the point $A(3, -4, 5)$ from the plane $2x + 5y - 6z = 16$ measured along the line $\frac{x}{2} = \frac{y}{1} = \frac{z}{-2}$ is