$A$ line joining the points $(1, 2, 0)$ and $(4, 13, 5)$ is perpendicular to a plane. Then the coefficients of $x, y$ and $z$ in the equation of the plane are respectively

Find the vector equation of the plane which is at a distance of $\frac{6}{\sqrt{29}}$ from the origin and its normal vector from the origin is $2 \hat{i}-3 \hat{j}+4 \hat{k} .$ Also find its cartesian form.

If the distance between the plane,$23x - 10y - 2z + 48 = 0$ and the plane containing the lines $\frac{x+1}{2} = \frac{y-3}{4} = \frac{z+1}{3}$ and $\frac{x+3}{2} = \frac{y+2}{6} = \frac{z-1}{\lambda}$ $(\lambda \in R)$ is equal to $\frac{k}{\sqrt{633}}$,then $k$ is equal to

The perpendicular distance of the origin from the plane $2x + y - 2z - 18 = 0$ is (in $\text{ units}$)

If the planes $\bar{r} \cdot(2 \hat{i}-\lambda \hat{j}+\hat{k})=3$ and $\bar{r} \cdot(4 \hat{i}-\hat{j}+\mu \hat{k})=5$ are parallel,then $\lambda+\mu=$

The coordinates of the foot of the perpendicular drawn from the origin to a plane is $(2, 4, -3)$. The equation of the plane is