Find the equation of the plane that passes through the three points $(1, 1, -1)$,$(6, 4, -5)$,and $(-4, -2, 3)$.

Let $Q$ be the foot of the perpendicular from the origin to the plane $4x - 3y + z + 13 = 0$ and $R$ be a point $(-1, 1, -6)$ on the plane. Then the length $QR$ is

$A$ plane passes through a fixed point $A(1, -2, 3)$. If the locus of the foot of the perpendicular to it from the point $B(0, 1, 2)$ is $x^2 + y^2 + z^2 + \alpha x + \beta y + \gamma z + \delta = 0$,then the value of $\alpha + \beta + \gamma + \delta$ is

The equation of the plane perpendicular to the line $\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$ and passing through the point $(2, 3, 4)$ is:

$A$ plane $x$ passes through the point $(1, 1, 1)$. If $b, c, a$ are the direction ratios of a normal to the plane,where $a, b, c$ $(a < b < c)$ are the factors of $2001$,then the equation of the plane is

The plane $\ell x+my=0$ is rotated about its line of intersection with the plane $z=0$ through an angle $\alpha$. The equation of the new plane is