If you decompose $M=\begin{pmatrix} X_{q\times q}&Y_{q\times k_3}\\ (Y_{q\times k_3})^{\rm T}&0_{k_3\times k_3}\end{pmatrix}$ into four block matrices, with $q=k_1+k_2$, then the determinant equals
$$\det M=(-1)^{k_3}(\det X_{q\times q})\det[(Y_{q\times k_3})^{\rm T}X_{q\times q}^{-1}Y_{q\times k_3}].$$
The second determinant has a root of multiplicity $k_3-q=k_3-k_1-k_2$.