Question:
Let $p_1$, $p_2$, $p_3$ be scalars and $V_1$, $V_2$, $V_3$ be $n \times n$ matrices. Further, the matrices can be expressed as
$$ V_1 = I_{d_2 d_3} \otimes W_1 $$ $$ V_2 = I_{d_3} \otimes W_2 \otimes I_{d_1} $$ $$ V_3 = W_3 \otimes I_{d_1 d_2} $$
where $W_i$ is a $d_i \times d_i$ matrix and $d_1 d_2 d_3 = n$. Then, what does the determinant of the following sum evaluate to?
$$ det(I_n - p_1 V_1 - p_2 V_2 - p_3 V_3) $$
Some notes / results I found
I know that there is the result
$$ det(\lambda I - A) = \prod^n_{i=1} (\lambda - \lambda_i) $$
where $\lambda_i$ are the eigenvalues of a matrix $A$. Hence, $ det(I-pA) = \prod^n_{i=1} (1-p\lambda_i) $. Hence, this problem can be reformulated to find the eigenvalues of the following sum
$$ p_1 V_1 + p_2 V_2 + p_3 V_3 $$
Also, the eigenvalue of a kronecker product is
$$ eig(A \otimes B) = \{ \lambda_i \omega_j : 1 \le i \le p, i \le j \le q \} $$
where $eig(A) = \{ \lambda_1, \ldots, \lambda_p \}$ and $eig(B) = \{ \omega_1, \ldots, \omega_q \}$ for $p\times p$ and $q \times q$ square matrices $A$ and $B$, respectively.