Let $$A = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$$ and $$B = \begin{pmatrix} 9^2 & -10^2 & 11^2 \\ 12^2 & 13^2 & -14^2 \\ -15^2 & 16^2 & 17^2 \end{pmatrix}$$, then the value of $$A'BA$$ is

We need to find the value of $$A'BA$$ where $$A = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$$ and $$B = \begin{pmatrix} 9^2 & -10^2 & 11^2 \\ 12^2 & 13^2 & -14^2 \\ -15^2 & 16^2 & 17^2 \end{pmatrix}$$.

$$A$$ is $$3 \times 1$$, so $$A'$$ (transpose) is $$1 \times 3$$.

$$B$$ is $$3 \times 3$$.

$$A'BA$$ is $$(1 \times 3)(3 \times 3)(3 \times 1) = 1 \times 1$$ (a scalar).

$$BA = \begin{pmatrix} 81 & -100 & 121 \\ 144 & 169 & -196 \\ -225 & 256 & 289 \end{pmatrix} \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$$

Row 1: $$81 - 100 + 121 = 102$$

Row 2: $$144 + 169 - 196 = 117$$

Row 3: $$-225 + 256 + 289 = 320$$

$$BA = \begin{pmatrix} 102 \\ 117 \\ 320 \end{pmatrix}$$

$$A'BA = \begin{pmatrix} 1 & 1 & 1 \end{pmatrix} \begin{pmatrix} 102 \\ 117 \\ 320 \end{pmatrix} = 102 + 117 + 320 = 539$$

Therefore, the correct answer is Option D: $$539$$.