Sort of positive matrix
If the blocks have dimension $1$, then, the sole solutions are $A=B=C\geq 0$.
EDIT. That follows is a solution that uses , like that of @user1551, the above remark.
We consider the matrix $M=U\otimes C+V\otimes B+W\otimes A$, where $U=\begin{pmatrix} 1&0&1\\ 0&0&0\\ 1&0&0\end{pmatrix},\cdots$. Then $(x\otimes y)^TM(x\otimes y)\geq 0$ for every vectors $x\in\mathbb{C}^3,y\in\mathbb{C}^k$, that is,
$x^TUx(y^TCy)+x^TVx(y^TBy)+x^TWx(y^TAy)\geq 0$; thus, for every $x$ and for a fixed $y$,
$x^T(uU+vV+wW)x\geq 0$, where $u=y^TCy,\cdots$. The associated matrix
$\begin{pmatrix}u&v&u\\v&v&w\\u&w&w\end{pmatrix}$ must be $\geq 0$, that happens only if $u=v=w$.
Finally, for every $y$, $y^TCy=y^TBy=y^TAy$, that implies that $A=B=C$.