Step 1: Project onto the normal:
The altitude is the component of $\vec c$ along the unit normal to the base, $\hat n = \frac{\vec a\times\vec b}{|\vec a\times\vec b|}$.
Step 2: Compute:
$\vec a\times\vec b = (-5, 3, 2)$, $|\cdot| = \sqrt{38}$. $h = |\vec c\cdot\hat n| = \frac{|-5 + 3 + 6|}{\sqrt{38}} = \frac{4}{\sqrt{38}} = \frac{2\sqrt2}{\sqrt{19}}$. Option (A).
Final Answer:
Option (A).
\[ \boxed{\frac{2\sqrt{2}}{\sqrt{19}}} \]