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

$A$ point moves such that the sum of its distances from the points $(4, 0, 0)$ and $(-4, 0, 0)$ remains $10$. The locus of the point is

If $(2,3,-3)$ is one end of a diameter of the sphere $x^2+y^2+z^2-6x-12y-2z+20=0$,then the other end of the diameter is

$A$ spherical ball is kept at the corner of a rectangular room such that the ball touches two perpendicular walls and lies on the floor. If a point on the sphere is at distances of $9, 16, 25$ from the two walls and the floor,then a possible radius of the sphere is

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 $\lambda=$

$A$ plane passes through a fixed point $(p, q, r)$ and cuts the axes at $A, B, C$. Then the locus of the centre of the sphere $OABC$ is