If the foot of the perpendicular drawn from $(0,0,0)$ to a plane is $(1,2,3)$,then the equation of the plane is:

The equation of the plane passing through the point $(1, 2, 1)$ and perpendicular to the planes $x + 2y + 2z - 7 = 0$ and $3x + 3y + 2z - 5 = 0$ is:

The equation of the plane passing through $(1, -1, 2)$ and perpendicular to the planes $x + 2y - 2z = 4$ and $3x + 2y + z = 6$ is:

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 Cartesian equation of the plane whose vector equation is $\vec{r}=(1+\lambda-\mu) \hat{i}+(2-\lambda) \hat{j}+(3-2 \lambda+2 \mu) \hat{k}$,where $\lambda, \mu$ are scalars,is:

The equation of the plane passing through the points $(2, 3, 1)$ and $(4, -5, 3)$ and parallel to the $X$-axis is: