Find: $\lim_{x→1}\frac{⁡Γ(1+x)Γ(2+x)⁡Γ(3+x)-12}{x-1}$

Consider first $$A=\Gamma (x+1) \Gamma (x+2) \Gamma (x+3)$$ Take logarithms and use the expansion around $x=1$ $$\log (\Gamma (x+a))=\log (\Gamma (a+1))+(x-1) \psi ^{(0)}(a+1)+\frac{1}{2} (x-1)^2 \psi ^{(1)}(a+1)+O\left((x-1)^3\right)$$ which make $$\log(A)=\log (12)+\left(\frac{13}{3}-3 \gamma \right) (x-1)+\left(\frac{\pi ^2}{4}-\frac{65}{36}\right) (x-1)^2+O\left((x-1)^3\right)$$ $$A=e^{\log(A)}=12+12 \left(\frac{13}{3}-3 \gamma \right) (x-1)+6 \left(\left(\frac{13}{3}-3 \gamma \right)^2+2 \left(\frac{\pi ^2}{4}-\frac{65}{36}\right)\right) (x-1)^2+O\left((x-1)^3\right)$$ $$\frac{A-12}{x-1}=(52-36 \gamma )+\left(91+6 \gamma (9 \gamma -26)+3 \pi ^2\right) (x-1)+O\left((x-1)^2\right)$$ which shows the limit and how it is approached.

Using this truncated formula for $x=1.1$ would give $36.08$ while the exact value would be $36.68$ .

Edit

If we make the problem more general $$L_n=\lim_{x→1}\Big[\frac{\prod_{i=1}^n \Gamma(x+i)-\prod_{i=1}^n \Gamma(1+i)}{x-1}\Big]$$ we have the following results $$\left( \begin{array}{cc} n & L_n \\ 1 & 1-\gamma \\ 2 & 5-4 \gamma \\ 3 & 52-36 \gamma \\ 4 & 1848-1152 \gamma \\ 5 & 300672-172800 \gamma \\ 6 & 277447680-149299200 \gamma \\ 7 & 1723509964800-877879296000 \gamma \\ 8 & 83234996748288000-40452677959680000 \gamma \\ 9 & 35395284161160806400000-16514401250259763200000 \gamma \end{array} \right)$$ These sequences are not in $OEIS$.

However, it is possible to show that $$\color{blue}{L_n=G(n+2) \Big((n+1) \left(H_{n+1}-1\right)-n\,\gamma \Big)}$$ where appears the Barnes G-function.

I prefer to add another answer for the general case.

$$P_n(x)=\frac{\prod_{i=1}^n \Gamma(x+i)-\prod_{i=1}^n \Gamma(1+i)}{x-1}$$ write, using the Barnes G-function, $$P_n(x)=\frac{G(x+1)^{n-1} G(x+2)^{-n} G(n+x+1) \Gamma (x+1)^n-G(n+2) }{x-1 }$$ Now, using $$\frac d {dx} G(x+a)=G(x+a) \left((x+a-1)\, \psi (x+a)-(x+a)+\frac{1+\log (2 \pi )}{2} \right)$$ and L'Hospital rule or simply the derivative $P'_n(1)$ this leads to the limit $$L_n= G(n+2) \Big((n+1) \big(\psi (n+2)-1\big)+\gamma \Big)$$ Using the identity $$H_{n+1}=\psi (n+2)+\gamma$$ we have the general result $$\color{blue}{L_n=G(n+2) \Big((n+1) \left(H_{n+1}-1\right)-n\,\gamma \Big)}$$