Find the equation of the plane which passes through the points $(0,1,2)$ and $(-1,0,3)$ and is perpendicular to the plane $2x+3y+z=5$.

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

Let $N$ be the foot of the perpendicular from the point $P(1, -2, 3)$ on the line passing through the points $(4, 5, 8)$ and $(1, -7, 5)$. Then the distance of $N$ from the plane $2x - 2y + z + 5 = 0$ is $.......$.

Find the distance of the point $(2, 1, 0)$ from the plane $2x + y + 2z + 5 = 0$.

If the distance between the plane $Ax-2y+z=d$ and the plane containing the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}$ is $\sqrt{6}$ units,then $|d|$ is

The point$(s)$ on the line $\vec r = \hat i + \hat j + \hat k + t(\hat i + 3\hat j - \hat k)$ at a distance of $3 \ units$ from the plane $\vec r \cdot (\hat i + 2\hat j + 2\hat k) + 2 = 0$ are