If $$1^2 \cdot ({^{15} C_{1}}) + 2^2 \cdot ({^{15} C_{2}}) + 3^2 \cdot ({^{15} C_{3}}) + \ldots + 15^2 \cdot ({^{15} C_{15}}) = 2^m \cdot 3^n \cdot 5^k$$, where $$m, n, k \in \mathbb{N}$$, then $$m + n + k$$ is equal to :

The required sum is $$S = \displaystyle\sum_{r=1}^{15} r^{2}\,\binom{15}{r}$$.

Start from the binomial expansion: $$(1+x)^{n} = \displaystyle\sum_{r=0}^{n} \binom{n}{r}\,x^{r}$$.

Differentiating once with respect to $$x$$ gives
$$\frac{d}{dx}(1+x)^{n} = n(1+x)^{\,n-1} = \sum_{r=1}^{n} r\,\binom{n}{r}\,x^{\,r-1}$$.

Multiplying by $$x$$: $$n(1+x)^{\,n-1}\,x = \sum_{r=1}^{n} r\,\binom{n}{r}\,x^{\,r}$$.

Differentiating a second time:
$$\frac{d}{dx}\!\left[n(1+x)^{\,n-1}\,x\right] = n(n-1)(1+x)^{\,n-2}\,x + n(1+x)^{\,n-1} = \sum_{r=1}^{n} r^{2}\,\binom{n}{r}\,x^{\,r-1}$$.

Again multiply by $$x$$ to match the powers:
$$n(n-1)(1+x)^{\,n-2}\,x^{2} + n(1+x)^{\,n-1}\,x = \sum_{r=1}^{n} r^{2}\,\binom{n}{r}\,x^{\,r}$$.

Now put $$x = 1$$ (because $$2^{n} = (1+1)^{n}$$):
Left side becomes $$n(n-1)\,2^{\,n-2}\,(1)^{2} + n\,2^{\,n-1}\,(1)$$.
Hence
$$\sum_{r=1}^{n} r^{2}\,\binom{n}{r} = n(n-1)\,2^{\,n-2} + n\,2^{\,n-1}$$.

Simplify the right side:
$$n(n-1)\,2^{\,n-2} + n\,2^{\,n-1} = n\,2^{\,n-2}\,\big[(n-1) + 2\big] = n(n+1)\,2^{\,n-2}$$ $$-(1)$$.

For our problem, $$n = 15$$. Substituting in $$(1)$$:
$$S = 15\,(15+1)\,2^{\,15-2} = 15 \times 16 \times 2^{\,13}$$.

Compute the numerical coefficients:
$$15 \times 16 = 240 = 2^{4}\times 3 \times 5$$.

Therefore
$$S = 240 \times 2^{\,13} = \bigl(2^{4}\times 3 \times 5\bigr)\,2^{\,13} = 2^{\,17}\,3^{\,1}\,5^{\,1}$$.

The required exponents are $$m = 17,\; n = 1,\; k = 1$$.

Hence $$m + n + k = 17 + 1 + 1 = 19$$.

So the correct option is Option A (19).