Largest value of determinant 
If $\alpha,\beta,\gamma \in [-3,10].$ Then largest value of the determinant
$$\begin{vmatrix}3\alpha^2&\beta^2+\alpha\beta+\alpha^2&\gamma^2+\alpha\gamma+\alpha^2\\\\
\alpha^2+\alpha\beta+\beta^2& 3\beta^2&\gamma^2+\beta\gamma+\beta^2\\\\
\alpha^2+\alpha\gamma+\gamma^2& \beta^2+\beta\gamma+\gamma^2&3\gamma^2\end{vmatrix}$$

Try: I am trying to break that determinant into product of 2 determinants but not able to break it.
Could someone help me in this question? Thanks.

Being a $3 \times 3$ determinant, it is too small to be attacked with sophisticated techniques, but also too involved to be attacked by brute force. We shall compute it with successive simplifications of its rows and columns. In the following, $R_i$ and $C_j$ will mean the $i$-th row and the $j$-th column, and $[R_i \to aR_i + bR_j]$ means to replace the $i$-th row with the linear combination $aR_i + bR_j$ where $a$ and $b$ are numbers. The computation, then, goes as follows:
$$\begin{align*}
& \begin{vmatrix}
3\alpha^2 & \beta^2 + \alpha \beta + \alpha^2 & \gamma^2 + \alpha \gamma + \alpha^2 \\
\alpha^2 + \alpha \beta + \beta^2 & 3\beta^2 & \gamma^2 + \beta \gamma + \beta^2 \\
\alpha^2 + \alpha \gamma + \gamma^2 & \beta^2 + \beta \gamma + \gamma^2 & 3\gamma^2\end{vmatrix} = [R_2 \to R_2 - R_3] = \\[10pt]
& \begin{vmatrix}
3\alpha^2 & \beta^2 + \alpha \beta + \alpha^2 & \gamma^2 + \alpha \gamma + \alpha^2 \\
(\beta - \gamma) (\alpha + \beta + \gamma) & (\beta - \gamma) (2\beta + \gamma) & (\beta - \gamma) (\beta + 2\gamma) \\
\alpha^2 + \alpha \gamma + \gamma^2 & \beta^2 + \beta \gamma + \gamma^2 & 3\gamma^2\end{vmatrix} = \\[10pt]
(\beta - \gamma) & \begin{vmatrix}
3\alpha^2 & \beta^2 + \alpha \beta + \alpha^2 & \gamma^2 + \alpha \gamma + \alpha^2 \\
\alpha + \beta + \gamma & 2\beta + \gamma & \beta + 2\gamma \\
\alpha^2 + \alpha \gamma + \gamma^2 & \beta^2 + \beta \gamma + \gamma^2 & 3\gamma^2\end{vmatrix} = [R_3 \to R_3 - R_1] = \\[10pt]
(\beta - \gamma) & \begin{vmatrix}
3\alpha^2 & \beta^2 + \alpha \beta + \alpha^2 & \gamma^2 + \alpha \gamma + \alpha^2 \\
\alpha + \beta + \gamma & 2\beta + \gamma & \beta + 2\gamma \\
(\gamma - \alpha) (\gamma + 2\alpha) & (\gamma - \alpha) (\alpha + \beta + \gamma) & (\gamma - \alpha) (2\gamma + \alpha) \end{vmatrix} = \\[10pt]
(\beta - \gamma) (\gamma - \alpha) & \begin{vmatrix}
3\alpha^2 & \beta^2 + \alpha \beta + \alpha^2 & \gamma^2 + \alpha \gamma + \alpha^2 \\
\alpha + \beta + \gamma & 2\beta + \gamma & \beta + 2\gamma \\
\gamma + 2\alpha & \alpha + \beta + \gamma & 2\gamma + \alpha \end{vmatrix} = [C_2 \to C_2 - C_1] = \\[10pt]
(\beta - \gamma) (\gamma - \alpha) & \begin{vmatrix}
3\alpha^2 & (\beta - \alpha) (\beta + 2\alpha) & \gamma^2 + \alpha \gamma + \alpha^2 \\
\alpha + \beta + \gamma & \beta - \alpha & \beta + 2\gamma \\
\gamma + 2\alpha & \beta - \alpha & 2\gamma + \alpha \end{vmatrix} = \\[10pt]
(\beta - \gamma) (\gamma - \alpha) (\beta - \alpha) & \begin{vmatrix}
3\alpha^2 & \beta + 2\alpha & \gamma^2 + \alpha \gamma + \alpha^2 \\
\alpha + \beta + \gamma & 1 & \beta + 2\gamma \\
\gamma + 2\alpha & 1 & 2\gamma + \alpha
\end{vmatrix} = [C_3 \to C_3 - C_1] \\[10pt]
(\beta - \gamma) (\gamma - \alpha) (\beta - \alpha) & \begin{vmatrix}
3\alpha^2 & \beta + 2\alpha & (\gamma - \alpha) (\gamma + 2\alpha) \\
\alpha + \beta + \gamma & 1 & \gamma - \alpha \\
\gamma + 2\alpha & 1 & \gamma - \alpha
\end{vmatrix} = \\[10pt]
(\beta - \gamma) (\gamma - \alpha)^2 (\beta - \alpha) & \begin{vmatrix}
3\alpha^2 & \beta + 2\alpha & \gamma + 2\alpha \\
\alpha + \beta + \gamma & 1 & 1 \\
\gamma + 2\alpha & 1 & 1
\end{vmatrix} = [C_3 \to C_3 - C_2] \\[10pt]
(\beta - \gamma) (\gamma - \alpha)^2 (\beta - \alpha) & \begin{vmatrix}
3\alpha^2 & \beta + 2\alpha & \gamma - \beta \\
\alpha + \beta + \gamma & 1 & 0 \\
\gamma + 2\alpha & 1 & 0
\end{vmatrix} = \\[10pt]
- (\beta - \gamma)^2 (\gamma - \alpha)^2 (\beta - \alpha) & \begin{vmatrix}
3\alpha^2 & \beta + 2\alpha & 1 \\
\alpha + \beta + \gamma & 1 & 0 \\
\gamma + 2\alpha & 1 & 0
\end{vmatrix} = \\[10pt]
- (\beta - \gamma)^2 (\gamma - \alpha)^2 (\beta - \alpha)^2
\end{align*}
$$
where the last $3 \times 3$ determinant has been expanded along the $3$rd column.
Since the determinant is the negative of the square of a product of real numbers, its maximum will be $0$, reached whenever any two of $\alpha, \beta, \gamma \in [-3,10]$ are equal.
