Evalute triple sum $ \sum_{m\geq 0}\sum_{n \geq 0}\sum_{p\geq 0} \frac{m!n!p!}{(m+n+p+2)!}$
Given the poles and residues of the $\Gamma$ function, or just by creative telescoping, we have $$\sum_{p\geq 0}\frac{p!}{(p+K)!}=\frac{1}{(K-1)\Gamma(K)}\tag{1} $$ hence $$\sum_{m,n,p\geq 0}\frac{m!n!p!}{(p+m+n+2)!}=\sum_{m,n\geq 0}\frac{\Gamma(m+1)\Gamma(n+1)}{(m+n+1)\Gamma(m+n+2)}=\sum_{m,n\geq 0}\frac{1}{(m+n+1)}\int_{0}^{1}x^n(1-x)^m\,dx\tag{2}$$ and by rearranging $$\sum_{m,n,p\geq 0}\frac{m!n!p!}{(p+m+n+2)!}=\int_{0}^{1}\sum_{m,n\geq 0}\frac{x^n(1-x)^m}{(m+n+1)}\,dx=\int_{0}^{1}\frac{\log(1-x)-\log(x)}{1-2x}\,dx\tag{3}$$ equals $$ 2\int_{0}^{1/2}-\log\left(\frac{x}{1-x}\right)\frac{dx}{1-2x} \stackrel{x\mapsto\frac{z}{1+z}}{=}2\int_{0}^{1}\frac{-\log(z)}{1-z^2}\,dz=2\sum_{n\geq 0}\frac{1}{(2n+1)^2}=\frac{\pi^2}{4}.\tag{4} $$
This approach can be generalized. For instance
$$\sum_{m,n,p\geq 0}\frac{m!n!p!}{(p+m+n+3)!}=\sum_{m,n\geq 0}\frac{\Gamma(m+1)\Gamma(n+1)}{(m+n+2)^2\Gamma(m+n+2)}=\sum_{m,n\geq 0}\frac{1}{(m+n+2)^2}\int_{0}^{1}x^n(1-x)^m\,dx\tag{2'}$$
and by rearranging
$$\sum_{m,n,p\geq 0}\frac{m!n!p!}{(p+m+n+3)!}=\int_{0}^{1}\sum_{m,n\geq 0}\frac{x^n(1-x)^m}{(m+n+2)^2}\,dx=\int_{0}^{1}\frac{x\text{Li}_2(1-x)-(1-x)\text{Li}_2(x)}{x(1-3x+2x^2)}\,dx\tag{3'}$$
equals
$$ 2\int_{0}^{1/2}\frac{x\text{Li}_2(1-x)-(1-x)\text{Li}_2(x)}{x(1-3x+2x^2)} \stackrel{x\mapsto\frac{z}{1+z}}{=}2\int_{0}^{1}\frac{z\text{Li}_2\left(\frac{1}{1+z}\right)-\text{Li}_2\left(\frac{z}{1+z}\right)}{z(1-z)}\,dz\tag{4'} $$
or
$$ 2\int_{0}^{1}\left[z\,\text{Li}_2\left(\frac{1}{1+z}\right)-\text{Li}_2\left(\frac{z}{1+z}\right)+(1-z)\,\text{Li}_2\left(\frac{1}{2-z}\right)-\text{Li}_2\left(\frac{1-z}{2-z}\right)\right]\frac{dz}{z}.$$
By integration by parts this expression boils down to Euler sums with weight $\leq 3$.
After a massive amount of computations we have
$$\boxed{\sum_{a,b,c\geq 0}\frac{a!b!c!}{(a+b+c+3)!}=\color{red}{\frac{13}{4}\zeta(3)-\frac{\pi^2}{2}\log(2)}.}$$