11 Find the largest real number $k$ such that for any positive real numbers $a, b, c, d$, the following inequality always holds: $(a+b+c)\left[3^{4}(a+b+c+d)^{5}+2^{4}(a+b+c+2 d)^{5}\right] \geqslant$ $k$ abcd ${ }^{3}$.
When $a=b=c=\frac{1}{3}, d=1$, we have $k \leqslant \frac{3^{4} \times 2^{5}+2^{4} \times 3^{5}}{\left(\frac{1}{3}\right)^{3}}=$ $2^{4} \times 3^{7} \times 5=174960$.

To prove the inequality $(a+b+c)\left[3^{4}(a+b+c+d)^{5}+2^{4}(a+b+c+\right.$ $\left.2 d)^{5}\right] \geqslant 174960 a b d^{3}$ for any $a, b, c, d \in \mathbf{R}^{*}$, let $a+b+c=t$, then $a b c \leqslant\left(\frac{a+b+c}{3}\right)^{3}=\frac{1}{27} t^{3}$. We need to prove
$$
\left.3^{4}(t+d)^{5}+2^{4}(t+2 d)^{5}\right] \geqslant 2^{4} \times 3^{4} \times 5 t^{3} d^{3},
$$

which is
$$
81(t+d)^{5}+16(t+2 d)^{5} \geqslant 6480 t^{2} d^{3}
$$

Since
$$
81(t+d)^{5}+16(t+2 d)^{5}
$$
$$
\begin{array}{l}
=97 t^{5}+565 t^{4} d+1450 t^{3} d^{2}+2090 t^{2} d^{3}+1685 t d^{4}+593 d^{5} \\
\geq(97+565+1450+2090+1685+593)\left(t^{\alpha} d^{\beta}\right)^{\frac{1}{188}},
\end{array}
$$

where
$$
\begin{aligned}
\alpha= & 97 \times 5+565 \times 4+1450 \times 3+2090 \times 2 \\
& +1685 \times 1=12960 . \\
\beta= & 565 \times 1+1450 \times 2+2090 \times 3+1685 \times 4 \\
& +593 \times 5=19440 .
\end{aligned}
$$

Therefore, $81(t+d)^{\overline{5}}+16(t+2 d)^{5} \geqslant 6480 t^{2} d^{3}$. In conclusion: $\cdot k_{\max }=174960$.