If $$\frac{^{n+2}C_6}{^{n-2}P_2} = 11$$, then $$n$$ satisfies the equation:

We have the relation

$$\frac{{}{}^{\,n+2}C_6}{{}^{\,n-2}P_2}=11.$$

First we recall the standard formulas:

Combination formula  $$\,{}^rC_k=\dfrac{r!}{k!(r-k)!}.$$

Permutation formula  $$\,{}^rP_k=\dfrac{r!}{(r-k)!}.$$

Using these, we write the two terms appearing in the given ratio.

For the numerator

$$ {}{}^{\,n+2}C_6=\frac{(n+2)!}{6!\,(n+2-6)!} =\frac{(n+2)!}{6!\,(n-4)!}. $$

For the denominator

$$ {}^{\,n-2}P_2=\frac{(n-2)!}{(n-2-2)!} =\frac{(n-2)!}{(n-4)!}. $$

Hence the given ratio becomes

$$ \frac{{}{}^{\,n+2}C_6}{{}^{\,n-2}P_2} = \frac{\dfrac{(n+2)!}{6!(n-4)!}} {\dfrac{(n-2)!}{(n-4)!}} = \frac{(n+2)!}{6!(n-2)!}. $$

Now we simplify the factorial quotient:

$$ \frac{(n+2)!}{(n-2)!} = (n+2)(n+1)n(n-1), $$

because the factors from $$(n-2)!$$ downward cancel completely inside $$(n+2)!$$.

So the ratio equals

$$ \frac{(n+2)(n+1)n(n-1)}{6!} = \frac{(n+2)(n+1)n(n-1)}{720}. $$

The problem states that this ratio is $$11$$, therefore

$$ \frac{(n+2)(n+1)n(n-1)}{720}=11. $$

Multiplying both sides by $$720$$ gives

$$ (n+2)(n+1)n(n-1)=7920, $$

because $$11\times720=7920$$.

The left‐hand side is the product of four consecutive integers. We can try successive integral values for $$n$$ near the fourth root of $$7920$$.

Checking $$n=9$$:

$$ (9+2)(9+1)\cdot9\cdot(9-1)=11\cdot10\cdot9\cdot8 =110\cdot72=7920. $$

Thus $$n=9$$ satisfies the equation.

Now we see which of the given quadratic equations becomes true when $$n=9$$.

Option A: $$n^2+n-110=9^2+9-110=81+9-110=-20\neq0.$$

Option B: $$n^2+2n-80=81+18-80=19\neq0.$$

Option C: $$n^2+3n-108=81+27-108=0.$$

Option D: $$n^2+5n-84=81+45-84=42\neq0.$$

Only Option C is satisfied.

Hence, the correct answer is Option C.