Implicitly Defined Surface

Let $\vec{P} = -5\hat{i} + 3\hat{j} + 9\hat{k}$ , $\vec{Q} = a\hat{i} + b\hat{j} + c\hat{k}$ , and $\vec{P} - \vec{Q} = (-5-a)\hat{i} + (3-b)\hat{j} + (9-c)\hat{k}$ .

Taking the above dot product gives:

$\vec{Q} \cdot (\vec{P} - \vec{Q}) = (a\hat{i} + b\hat{j} + c\hat{k}) \cdot [(-5-a)\hat{i} + (3-b)\hat{j} + (9-c)\hat{k}] = 0;$

or $(-5a-a^{2}) + (3b-b^{2}) + (9c-c^{2}) = 0;$

or $(\frac{25}{4} + 5a + a^{2}) + (\frac{9}{4} - 3b + b^{2}) + (\frac{81}{4} - 9c + c^{2}) = \frac{25}{4} + \frac{9}{4} + \frac{81}{4};$

or $(a + \frac{5}{2})^{2} + (b - \frac{3}{2})^{2} + (c - \frac{9}{2})^{2} = \frac{115}{4}.$

Hence, the surface is a sphere with a radius = $\frac{\sqrt{115}}{2}$ and a volume equal to $V = \frac{4\pi}{3} r^{3} = \frac{4\pi}{3} (\frac{\sqrt{115}}{2})^{3} = \boxed{\frac{\pi}{6} \cdot 115^{\frac{3}{2}}}.$

Thales' theorem: If $A,B$ lie diametrically across from each other on a circle, then $C$ is a point on that circle iff $\angle ACB = 90^\circ$ .

The condition in this problem may be written $(Q-O)\cdot(Q-P) = 0$ , where $O$ is the origin; in other words, $\angle OQP = 90^\circ$ ; which means that $Q$ can be any point on any circle of which $OP$ is a diameter. This means that $Q$ traces out a sphere, with $OP$ as a diameter.

The Pythagorean Theorem shows that $d = |OP| = \sqrt{115}$ , so $r = \tfrac12\sqrt{115}$ . Plug this into $V = \frac{4\pi}3 r^3$ to find the final answer, $\boxed{645.722}$ .