12 Find the maximum constant $k$, such that $\frac{k a b c}{a+b+c} \leqslant(a+b)^{2}+(a+b+4 c)^{2}$ holds for all positive real numbers $a$, $b$, $c$.
12. Let $a=b=2c$, then $k \leqslant 100$. Also, due to
$$\begin{aligned}
& \frac{a+b+c}{a b c} \cdot\left[(a+b)^{2}+(a+b+4 c)^{2}\right] \\
\geqslant & \frac{a+b+c}{a b c} \cdot\left[(a+b)^{2}+(a+2 c+b+2 c)^{2}\right] \\
\geqslant & \frac{a+b+c}{a b c} \cdot\left[4 a b+(2 \sqrt{2 a c}+2 \sqrt{2 b c})^{2}\right] \\
= & \frac{a+b+c}{a b c} \cdot(4 a b+8 a c+8 b c+16 \sqrt{a c} \cdot c) \\
= & (a+b+c)\left(\frac{4}{c}+\frac{8}{a}+\frac{8}{b}+\frac{16}{\sqrt{a b}}\right) \\
= & \left(\frac{a}{2}+\frac{a}{2}+\frac{b}{2}+\frac{b}{2}+c\right) \cdot\left(\frac{4}{c}+\frac{8}{a}+\frac{8}{b}+\frac{8}{\sqrt{a b}}+\frac{8}{\sqrt{a b}}\right) \\
\geqslant & \left(5 \sqrt[5]{\frac{a^{2} b^{2} c}{2^{4}}}\right) \cdot\left(5 \sqrt[5]{\frac{2^{2}}{a^{2} b^{2} c}}\right)=100 .
\end{aligned}$$

Thus, $k_{\max }=100$.