If $$A = \left[\begin{array}{ccc} -6 &1 &-6\cr 1 &5 &9\cr -6 &9 &1 \end{array}\right]$$ and $Q(\vec{x})=\vec{x}^T A \vec{x}$, then

$Q(x_1, x_2, x_3)=$
Enter $x_1$ as x1, $x_2$ as x2, and $x_3$ as x3.