Given that, det $\begin{pmatrix}
2 &\alpha \\
3 & \beta
\end{pmatrix} = 16$
$\implies 2\beta - 3\alpha = 16 \quad \rightarrow (1)$
And, det $\begin{pmatrix}
2 &\beta \\
3 & \alpha
\end{pmatrix} = 1$
$\implies 2\alpha - 3\beta = 1\quad \rightarrow (2)$
On solving the equation $(1)$ and $(2),$ we get $\alpha = -10,\beta = -7.$
Now, $\alpha^{2} + \beta^{2} = (-10)^{2} + (-7)^{2} = 100 + 49 = 149.$
So, the correct answer is $149.$