The answer is 322.

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Because F(x) is monic we can say:

$F(x)={ x }^{ 3 }+a{ x }^{ 2 }+bx+c$

Define a quartic polynomial P(x) such that:

$P(x)=F(x)-x^{ 4 }\\ P(x)=-x^{ 4 }+{ x }^{ 3 }+a{ x }^{ 2 }+bx+c$

Note that 1, 2, and 3 are roots of P(x). Let us call the 4th unknown root r.By vieta's formulas

$1+2+3+r=-\frac { 1 }{ -1 } =1\\ r=-5$

Knowing its four roots we may rewrite P(x) as

$P(x)=-(x-1)(x-2)(x-3)(x+5)$

Now solve for P(4) and use it to solve for F(4)

$P(4)=-(3)(2)(1)(9)=-54\\ P(4)=F(4)-256\\ -54=F(4)-256\\ 202=F(4)$

Finally, calculate $\Gamma (6)$ and add it to F(4) to get the answer

$\Gamma (6)=(6-1)!=5!=120\\ F(4)+\Gamma (6)=202+120=322$