Answer
$\det A=-36$
Work Step by Step
Given: $A=\begin{vmatrix}
1 & 2 & 3\\
2 & 6 & 4\\
3 & -5 &2
\end{vmatrix}$
So we get
$\begin{vmatrix}
1 & 2 & 3\\
2 & 6 & 4\\
3 & -5 &2
\end{vmatrix} \xrightarrow{R_2-2R_1,R_3-3R_1} \begin{vmatrix}
1 & 2 & 3\\
0 & 2 & -2\\
0 &-11 &-7
\end{vmatrix} \xrightarrow{R_3+\frac{11}{2}R_2} \begin{vmatrix}
1 & 2 & 3\\
0 & 2 & -2\\
0 & 0 &-18
\end{vmatrix}$
Determinant of triangular matrix is the product of its diagonal elements, so: $\det A=1⋅2⋅(−18)=−36$
