Answer
$\det (B)=24$
Work Step by Step
Let $A=\begin{bmatrix}
a & b & c\\
d & e &f\\
g & h & i
\end{bmatrix}$ and $\det (A)=6$
Permuting the first and second rows of $A$, and then the second and third rows of $A$, we obtain
$A=\begin{bmatrix}
d & e &f\\
g & h & i\\
a & b & c
\end{bmatrix}$
$\det (A_1)=-6$
Multiplying the first row of $A_1$ by $-4$, we obtain:
$A_2=\begin{bmatrix}
-4d & -4e &-4f\\
g & h & i\\
a & b & c
\end{bmatrix}$
$\det (A_2)=(-4)\det (A_1)=-4.(-6)=24$
Adding 5 times third row of $A_2$ to the second row, we obtain:
$B=\begin{bmatrix}
-4d & -4e &-4f\\
g+5a & h+5b & i+5c\\
a & b & c
\end{bmatrix}$
$\det (B)=\det (A_2)=24$
