We know the $n^{th}$ term of an A.P. is given by,

$a_n=a+\left(n-1\right)d$

Given, $a_7=3$

$\Rightarrow a+6d=3$

$\Rightarrow a=3-6d$

And, $a_1{a}_4=a\left(a+3d\right)$

$=\left(3-6d\right)\left(3-3d\right)$

$=18d^2-27d+9$

Given product $\left(a_1{a}_4\right)$ is minimum then,

Let $f\left(d\right)=18d^2-27d+9$

$f^{\text{'}}\left(d\right)=36d-27$

Product to be minimum, $f^{\text{'}}\left(d\right)=0$

$\Rightarrow 36d-27=0$

$\Rightarrow d=\frac{27}{36}=\frac{3}{4}$

So, $a=3-\frac{9}{2}=\frac{-3}{2}$

Given, $S_n=0$

$S_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right]=0$

$-3+\left(n-1\right)\frac{3}{4}=0$

$\Rightarrow n=5$

Now $n!-4a_{n(n+2)}=5!-4a_{35}$

$=120-4\left(a+34d\right)$

$=120-4\left(\frac{-3}{2}+34\times \frac{3}{4}\right)$

$=120+6-102=24$