Let $$a_1, a_2, a_3, \ldots$$ be an A.P. If $$a_7 = 3$$, the product $$(a_1 a_4)$$ is minimum and the sum of its first $$n$$ terms is zero then $$n! - 4a_{n(n+2)}$$ is equal to

The terms of an A.P. are $$a_1 , a_2 , a_3 ,\ldots$$ with common difference $$d$$, so $$a_k = a_1 + (k-1)d$$.

Given $$a_7 = 3$$:
$$a_7 = a_1 + 6d = 3 \;\; -(1)$$

The product $$P = a_1 a_4$$ must be minimum.
Since $$a_4 = a_1 + 3d$$, we have $$P = a_1(a_1 + 3d) = a_1^{\,2} + 3d a_1.$$ From (1), $$a_1 = 3 - 6d$$, so $$P(d) = (3 - 6d)^2 + 3d(3 - 6d) = 9 - 36d + 36d^2 + 9d - 18d^2 = 18d^2 - 27d + 9.$$

For a quadratic $$ad^2 + bd + c$$, the minimum occurs at $$d = -\dfrac{b}{2a}.$$
Thus $$d = -\dfrac{-27}{2 \times 18} = \dfrac{27}{36} = \dfrac{3}{4}.$$

Substituting $$d = \dfrac{3}{4}$$ into (1):
$$a_1 = 3 - 6\left(\dfrac{3}{4}\right) = 3 - \dfrac{9}{2} = -\dfrac{3}{2}.$$

The sum of the first $$n$$ terms is zero.
For an A.P., $$S_n = \dfrac{n}{2}\bigl(2a_1 + (n-1)d\bigr).$$
Therefore $$\dfrac{n}{2}\Bigl(-3 + (n-1)\dfrac{3}{4}\Bigr)=0.$$ Since $$n \neq 0$$, the parenthesis must vanish: $$-3 + (n-1)\dfrac{3}{4}=0 \;\Rightarrow\; (n-1)\dfrac{3}{4}=3 \;\Rightarrow\; n-1 = 4 \;\Rightarrow\; n = 5.$$

We need $$n! - 4a_{n(n+2)}.$$
Here $$n(n+2)=5 \times 7 = 35$$, so $$a_{35} = a_1 + 34d = -\dfrac{3}{2} + 34\left(\dfrac{3}{4}\right) = -\dfrac{3}{2} + \dfrac{102}{4} = -\dfrac{3}{2} + \dfrac{51}{2} = \dfrac{48}{2} = 24.$$

Now evaluate: $$n! - 4a_{35} = 5! - 4 \times 24 = 120 - 96 = 24.$$

Option D which is: 24