The centre of the sphere $\alpha \,r^2 - 2u \cdot r = \beta ,(\alpha \ne 0)$ is

The equation of the sphere passing through the points $(1,0,0), (0,1,0)$ and $(1,1,1)$ and having the smallest radius is:

The shortest distance from the origin to a variable point on the sphere $(x-2)^2+(y-3)^2+(z-6)^2=1$ is

The coordinate of a point equidistant from the points $(0, 0, 0), (a, 0, 0), (0, b, 0),$ and $(0, 0, c)$ is

The plane $x + 2y - z = 4$ cuts the sphere $x^2 + y^2 + z^2 - x + z - 2 = 0$ in a circle of radius:

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