If $A(3, -2, 2)$ and $B(2, \lambda+1, 5)$ are the end points of the diameter of a circle and the point $P(5, 6, -1)$ lies on the circle,then $\lambda=$

Let $(3, 4, -1)$ and $(-1, 2, 3)$ be the end points of a diameter of a sphere. Then the radius of the sphere is equal to

What is the center of the sphere passing through the four points $(0, 0, 0), (0, 2, 0), (1, 0, 0),$ and $(0, 0, 4)$?

If the plane $2ax - 3ay + 4az + 6 = 0$ passes through the midpoint of the line joining the centres of the spheres ${x^2} + {y^2} + {z^2} + 6x - 8y - 2z = 13$ and ${x^2} + {y^2} + {z^2} - 10x + 4y - 2z = 8$,then $a$ equals

$A$ variable plane passes through a fixed point $P(1, 2, 3)$. The foot of the perpendicular from the origin $O(0, 0, 0)$ to the plane lies on:

The ratio in which the sphere ${x^2} + {y^2} + {z^2} = 504$ divides the line segment $AB$ joining the points $A(12, -4, 8)$ and $B(27, -9, 18)$ is given by