$\textbf{A faster route: check if the Hessian even depends on the point.}$
The function is $f(x,y) = x^2 + y^2 + xy - 8x - 7y$. Every term is either quadratic or linear in $x$ and $y$.
The second partial derivatives of a linear term are always zero, and the second partial derivatives of a quadratic term are constants. So the Hessian of this $f$ is the same matrix everywhere in the plane, we do not have to find the stationary point first to answer this question.
$\textbf{Get the second derivatives directly.}$
$f_x = 2x + y - 8$, so $f_{xx} = 2$ and $f_{xy} = 1$.
$f_y = x + 2y - 7$, so $f_{yy} = 2$ and $f_{yx} = 1$.
These match, confirming $f_{xy} = f_{yx}$ as expected for a smooth function.
$\textbf{Build the Hessian and take its determinant.}$
\[ H = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix} \]
$\det(H) = (2)(2) - (1)(1) = 3$.
For the record, the stationary point itself is $(3, 2)$, found by solving $2x+y=8$ and $x+2y=7$ together, but that step was not even needed to answer this question.
$\textbf{Final answer.}$
The determinant of the Hessian is 3, which is option (B).
\[ \boxed{3} \]