A rectangular box measures $a \times b \times c$, where $a$, $b$, and $c$ are integers and $1\leq a \leq b \leq c$. The volume and the surface area of the box are numerically equal. How many ordered triples $(a,b,c)$ are possible?
$\textbf{(A)}\; 4 \qquad\textbf{(B)}\; 10 \qquad\textbf{(C)}\; 12 \qquad\textbf{(D)}\; 21 \qquad\textbf{(E)}\; 26$
We need \[abc = 2(ab+bc+ac) \quad \text{ or } \quad (a-2)bc = 2a(b+c).\]
Since $a\le b, ac \le bc$, from the first equation we get $abc \le 6bc$. Thus $a\le 6$. From the second equation we see that $a > 2$. Thus $a\in \{3, 4, 5, 6\}$. 

If $a=3$ we need $bc = 6(b+c) \Rightarrow (b-6)(c-6)=36$. We get five roots $\{(3, 7, 42), (3, 8, 24), (3,9,18), (3, 10, 15), (3,12,12)\}$.
If $a=4$ we need $2bc = 8(b+c) \Rightarrow bc = 4(b+c) \Rightarrow (b-4)(c-4)=16$. We get three roots $\{(4,5,20), (4,6,12), (4,8,8)\}$.
If $a=5$ we need $3bc = 10(b+c) \Rightarrow 9bc=30(b+c) \Rightarrow (3b-10)(3c-10)=100$. We get one root $\{(5,5,10)\}$.
If $a=6$ we need $4bc = 12(b+c) \Rightarrow bc = 3(b+c) \Rightarrow (b-3)(c-3)=9$. We get  one root $\{(6,6,6)\}$.
Thus, there are $5+3+1+1 = \boxed{\textbf{(B)}\; 10}$ solutions.