Answer
$\det (B)=-12$
Work Step by Step
Let $A=\begin{bmatrix}
a & b & c\\
d & e &f\\
g & h & i
\end{bmatrix}$ and $\det (A)=-6$
$A^T=\begin{bmatrix}
a & d & g\\
b & e &h\\
c& f & i
\end{bmatrix}$
$\det (A_1)=\det (A^T)=-6$
Adding $-1$ times third row of $A^T$ to the second row, we obtain
$A_2=\begin{bmatrix}
a & d & g\\
b-c & e-f &h-i\\
c& f & i
\end{bmatrix}$
$\det (A_2)=\det (A^T)=-6$
Multiplying first row of $A_2$ by $-1$ to the third row, we obtain:
$A_3=\begin{bmatrix}
a & d & g\\
b-c & e-f &h-i\\
c-a& f-d & i-g
\end{bmatrix}$
$\det (A_3)=\det (A_2)=-6$
Multiplying first row of $A_3$ by $2$:
$B=\begin{bmatrix}
2a & 2d & 2g\\
b-c & e-f &h-i\\
c-a& f-d & i-g
\end{bmatrix}$
$\det (B)=2\det (A_3)=-12$
