Step1:
Given
$A=\begin{bmatrix}1 & 5\\7 & 12\end{bmatrix}.$
$B=\begin{bmatrix}9 & 1\\7 & 8\end{bmatrix}$
3A+5B+2C=0
Substitute the value of A and B in the above given equation.
$3\begin{bmatrix}1 & 5\\7 & 12\end{bmatrix}+5\begin{bmatrix}9 & 1\\7 & 8\end{bmatrix}+2C=0.$
$\Rightarrow \begin{bmatrix}3 & 15\\21 & 36\end{bmatrix}+\begin{bmatrix}45 & 5\\35 & 40\end{bmatrix}+2C=0.$
$\Rightarrow \begin{bmatrix}3+45 & 15+5\\21+35 & 36+40\end{bmatrix}+2C=0.$
$\Rightarrow \begin{bmatrix}48 & 20\\56 & 76\end{bmatrix}+2C=0.$
Step2:
$\Rightarrow 2C=(-)\begin{bmatrix}48 & 20\\56 & 76\end{bmatrix}.$
$C=-1/2\begin{bmatrix}48 & 20\\56 & 76\end{bmatrix}$
$\;\;\;=\begin{bmatrix}-24 & -10\\-28 & -38\end{bmatrix}$
$C=\begin{bmatrix}-24 & -10\\-28 & -38\end{bmatrix}$