Triangle $ABC$, with sides of length $5$, $6$, and $7$, has one [vertex](https://artofproblemsolving.com/wiki/index.php/Vertex) on the positive $x$-axis, one on the positive $y$-axis, and one on the positive $z$-axis. Let $O$ be the [origin](https://artofproblemsolving.com/wiki/index.php/Origin). What is the volume of [tetrahedron](https://artofproblemsolving.com/wiki/index.php/Tetrahedron) $OABC$?
$\mathrm{(A)}\ \sqrt{85}\qquad\mathrm{(B)}\ \sqrt{90}\qquad\mathrm{(C)}\ \sqrt{95}\qquad\mathrm{(D)}\ 10\qquad\mathrm{(E)}\ \sqrt{105}$
Without loss of generality, let $A$ be on the $x$ axis, $B$ be on the $y$ axis, and $C$ be on the $z$ axis, and let $AB, BC, CA$ have respective lengths of 5, 6, and 7.  Let $a,b,c$ denote the lengths of segments $OA,OB,OC,$ respectively.  Then by the [Pythagorean Theorem](https://artofproblemsolving.com/wiki/index.php/Pythagorean_Theorem),
\begin{align*} a^2+b^2 &=5^2 , \\  b^2+c^2&=6^2, \\ c^2+a^2 &=7^2 , \end{align*}
so $a^2 = (5^2+7^2-6^2)/2 = 19$; similarly, $b^2 = 6$ and $c^2 = 30$. Since $OA$, $OB$, and $OC$ are mutually perpendicular, the tetrahedron's volume is \[abc/6\] because we can consider the tetrahedron to be a right triangular pyramid. 
\[abc/6 = \sqrt{a^2b^2c^2}/6 = \frac{\sqrt{19 \cdot 6 \cdot 30}}{6} = \sqrt{95},\]
which is answer choice $\boxed{\text{C}}$.