Let $$X = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$$ and $$A = \begin{pmatrix} -1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & -1 \end{pmatrix}$$. For $$k \in \mathbb{N}$$, if $$X'A^kX = 33$$, then $$k$$ is equal to

We have $$X = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$$ and $$A = \begin{pmatrix} -1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & -1 \end{pmatrix}$$, and we seek $$k \in \mathbb{N}$$ such that $$X'A^kX = 33$$.

Rather than computing $$A^k$$ in closed form, we observe that $$X'A^kX$$ can be computed iteratively since $$A^kX = A(A^{k-1}X)$$. Let us define $$V_k = A^kX$$ and track the sequence $$X'V_k$$.

We compute $$V_1 = AX = \begin{pmatrix} -1+2+3 \\ 0+1+6 \\ 0+0-1 \end{pmatrix} = \begin{pmatrix} 4 \\ 7 \\ -1 \end{pmatrix}$$, giving $$X'V_1 = 4 + 7 - 1 = 10$$.

Now $$V_2 = AV_1 = \begin{pmatrix} -4+14-3 \\ 0+7-6 \\ 0+0+1 \end{pmatrix} = \begin{pmatrix} 7 \\ 1 \\ 1 \end{pmatrix}$$, giving $$X'V_2 = 9$$.

Continuing, $$V_3 = AV_2 = \begin{pmatrix} -7+2+3 \\ 0+1+6 \\ 0+0-1 \end{pmatrix} = \begin{pmatrix} -2 \\ 7 \\ -1 \end{pmatrix}$$, giving $$X'V_3 = 4$$. Next, $$V_4 = AV_3 = \begin{pmatrix} 2+14-3 \\ 0+7-6 \\ 0+0+1 \end{pmatrix} = \begin{pmatrix} 13 \\ 1 \\ 1 \end{pmatrix}$$, giving $$X'V_4 = 15$$.

A clear pattern emerges. For even $$k$$, $$V_k$$ has the form $$\begin{pmatrix} c \\ 1 \\ 1 \end{pmatrix}$$, and for odd $$k$$, $$V_k = \begin{pmatrix} d \\ 7 \\ -1 \end{pmatrix}$$. At each even step, the first component increases by 6: $$c_2 = 7, c_4 = 13, c_6 = 19, \ldots$$, so $$c_{2m} = 6m + 1$$. This gives $$X'A^{2m}X = (6m + 1) + 1 + 1 = 6m + 3$$.

Setting $$6m + 3 = 33$$ yields $$m = 5$$, so $$k = 2m = 10$$. We can verify the odd-$$k$$ formula gives $$X'A^{2m+1}X = -6m + 10$$, which never equals 33 for any natural number $$m$$, confirming that $$k = 10$$ is the unique solution.

Hence, the correct answer is $$\boxed{10}$$.