Find the equation of the set of points which are equidistant from the points $A(3, 4, -5)$ and $B(-2, 1, 4)$.

The foot of the perpendicular drawn from the origin to the plane is $(4, -2, 5)$. Then,the Cartesian equation of the plane is:

The value of $k$ for which the planes $3x - 6y - 2z = 7$ and $2x + y - kz = 5$ are perpendicular to each other,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

$A$ plane $\pi$ passing through the point $3 \hat{i}-4 \hat{j}+5 \hat{k}$ is parallel to the plane which passes through the point $\hat{i}+\hat{j}-\hat{k}$ and is perpendicular to the vector $\hat{i}+2 \hat{j}-3 \hat{k}$. Then the Cartesian equation of $\pi$ is

Let $P(\lambda, 2, 1)$ be a point on the plane which passes through the point $Q(4, -2, 2)$. If the plane is perpendicular to the line joining the points $A(-2, -21, 29)$ and $B(-1, -16, 33)$,then find the value of $\left(\frac{\lambda}{11}\right)^{2} - \frac{4\lambda}{11} - 4$.