Let A be a 3×3 matrix of non-negative real elements such that $$A\begin{bmatrix}1\\1\\1\end{bmatrix} = 3\begin{bmatrix}1\\1\\1\end{bmatrix}$$. Then the maximum value of det(A) is ______.

Let $$A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}$$

Now

$$A \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = 3 \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$$

$$\begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 3 \\ 3 \\ 3 \end{bmatrix}$$

$$a_{11} + a_{12} + a_{13} = 3$$

$$a_{21} + a_{22} + a_{23} = 3$$

$$a_{31} + a_{32} + a_{33} = 3$$

Now for maximum value of $$\det(A)$$, choosing $$a_{ij} = \begin{cases} 0 & i \neq j \\ 3 & i = j \end{cases}$$

$$\therefore \vert{}A\vert{} = 27$$