Consider $A$ a $2 \times 3 $ matrix, $B$ a $3 \times 3 $ diagonal matrix with non-zero elements, and $C$ a $2 \times 2$ matrix that is also symmetric. The symmetric matrix $A A^T$ is invertible. Consider $\gamma \neq 0$.
Is there any way to solve the matrix equation $$ A B A^T - \gamma C=0 $$ for the matrix $B$?
Can I find conditions on $\gamma$ and $B$ such that $$ \det(A B A^T - \gamma C)=0?$$
So the main problem here is that the matrix $A$ is not square. I thought maybe there is a way to "complete" the matrix $A$ without messing the result of the determinant, however, I had no success with this idea.
$$ {\bf A} \operatorname{diag} ({\bf x}) \, {\bf A}^\top = {\bf B} $$
Let ${\bf a}_i$ denote the $i$-th column of $\bf A$. Hence,
$$ {\bf A} \operatorname{diag} ({\bf x}) \, {\bf A}^\top = x_1 {\bf a}_1 {\bf a}_1^\top + x_2 {\bf a}_2 {\bf a}_2^\top + x_3 {\bf a}_3 {\bf a}_3^\top = {\bf B} $$
Due to symmetry, half-vectorizing both sides yields a linear system of $3$ equations in $3$ unknowns
$$ \begin{bmatrix} | & | & | \\ \mbox{vech} \left( {\bf a}_1 {\bf a}_1^\top \right) & \mbox{vech} \left( {\bf a}_2 {\bf a}_2^\top \right) & \mbox{vech} \left( {\bf a}_3 {\bf a}_3^\top \right) \\ | & | & | \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} | \\ \mbox{vech} ({\bf B}) \\ | \end{bmatrix} $$