1. Write the A.P. in standard form

Let the A.P. have first term $a$ and common difference $d$. Then $a_r=a+(r-1)d$

Given: $a_7=a+6d=3 \,\,\,\text{(1)}$

Also, $a_1a_4=a(a+3d)$

We are told this product is minimum.

2. Use the condition for minimum product

From (1), $a=3-6d$

So, $a_1a_4=a(a+3d)=(3-6d)(3-3d)$

Expand: $(3-6d)(3-3d)=9-27d+18d^2$

This is a quadratic in $d$ with positive coefficient, so its minimum occurs at $d=\frac{-(-27)}{2\cdot 18}=\frac{27}{36}=\frac34$

Then $a=3-6\left(\frac34\right)=3-\frac{18}{4}=3-\frac92=-\frac32$

Hence the A.P. is $a_1=-\frac32,\qquad d=\frac34$

3. Use the condition that sum of first $n$ terms is zero

Sum of first $n$ terms of an A.P. is $S_n=\frac n2\left[2a+(n-1)d\right]$

Given $S_n=0$. Since $n\neq 0$, $2a+(n-1)d=0$

Substitute $a=-\frac32$ and $d=\frac34$: $2\left(-\frac32\right)+(n-1)\frac34=0$ $-3+\frac34(n-1)=0$ $\frac34(n-1)=3$ $n-1=4$ $n=5$

4. Find $a_{n(n+2)}$

First compute: $n(n+2)=5\cdot 7=35$

So we need $a_{35}$.

Using $a_r=a+(r-1)d$ $a_{35}=-\frac32+34\cdot \frac34$ $a_{35}=-\frac32+\frac{102}{4}=-\frac32+\frac{51}{2}=\frac{48}{2}=24$

Thus, $a_{n(n+2)}=24$

5. Evaluate the required expression

The printed expression is $n!-4a_{n(n+2)}$. With $n=5$, $5!-4a_{35}=120-4(24)=120-96=24$

So the value is $\boxed{24}$

6. Check with options

Option A is $24$.

So the correct answer is A.