6. Let the edge lengths of three regular tetrahedra be integers, and the sum of their volumes is $\frac{51}{4} \sqrt{2}$. Then the sum of the surface areas of these regular tetrahedra is $\qquad$
Reference answer: Let the edge lengths of the three regular tetrahedra be $a$, $b$, and $c$, with $0 < a \leq b \leq c$.
From $\frac{\sqrt{2}}{12}\left(a^{3}+b^{3}+c^{3}\right)=\frac{51}{4} \sqrt{2}$, we get $a^{3}+b^{3}+c^{3}=153$. Among these, $c^{3} \leq 125$, i.e., $c \leq 5$.
Since $3 c^{3} \geq a^{3}+b^{3}+c^{3}=153$, it follows that $c^{3} \geq 51$, so $c \geq 4$. Therefore, $c=4$ or $c=5$.
If $c=4$, there do not exist suitable $a, b$; if $c=5$, it is easy to find $a=1, b=3$. The sum of the surface areas is:
$$
4\left(\frac{1}{2} a^{2} \sin 60^{\circ}+\frac{1}{2} b^{2} \sin 60^{\circ}+\frac{1}{2} c^{2} \sin 60^{\circ}\right)=\sqrt{3}\left(a^{2}+b^{2}+c^{2}\right)=\sqrt{3}\left(1^{2}+3^{2}+5^{2}\right)=35 \sqrt{3} \text {. }
$$