If the foot of the perpendicular 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 $(1, 2, -3)$ and perpendicular to the planes $3x + y - 2z = 5$ and $2x - 5y - z = 7$ is:

The perpendicular distance from the origin to the plane containing the points having position vectors $\hat{i}+2\hat{j}+3\hat{k}$, $2\hat{i}+3\hat{j}-4\hat{k}$, and $3\hat{i}-4\hat{j}+5\hat{k}$ is

The plane,passing through the points $(0,-1,2)$ and $(-1,2,1)$ and parallel to the line passing through $(5,1,-7)$ and $(1,-1,-1)$,also passes through the point.

Find the equation of the plane that contains the point $(1, -1, 2)$ and is perpendicular to each of the planes $2x + 3y - 2z = 5$ and $x + 2y - 3z = 8$.

$P_1$ and $P_2$ are two distinct and intersecting planes. Three non-collinear points lie on $P_1$ and another three non-collinear points lie on $P_2$ (none being on the line of intersection of the planes). Then the maximum number of tetrahedrons formed using these six points is: