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 $k_3-q=k_3-k_1-k_2$.

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 $k_3-q=k_3-k_1-k_2$.

For $k_1\neq k_2$ the matrix $X$ is not invertible: We can give it an infinitesimal perturbation, $M\mapsto M_\epsilon=M+\epsilon 1_{q\times q}$, and then $\det (\lambda-M_\epsilon)=\lambda^{k_3-q}f_\epsilon(\lambda)$. The continuity of the determinant in the matrix elements ensures that the multiplicity of the root 0 cannot decrease in the limit $\epsilon\rightarrow 0$.