One should not solve the same problem for all sittings (one-trick pony?)
The same solution
a. $\frac{dW}{W}=\frac{dt}{0}\implies W=C$.
b. Characteristic equation $L(k):= k^3-7k+6= (k-1)(k-2)(k+3)=0\implies k_1=1$, $k_2=2$, $k_3=-3$. Then
$y_1=e^t$, $y_2=e^{2t}$, $y_3=e^{-3t}$ and
$$
W(y_1,y_2, y_3)=\left|\begin{matrix}
e^t &e^{2t} &e^{-3t}\\
e^t &2e^{2t} &-3e^{-t}\\
e^t &4e^{2t} &9e^{-2t}
\end{matrix}\right|\overset{(A)}{=}
\left|\begin{matrix}
1 &1 &1\\
1 &2 &-3\\
1 &4 &9
\end{matrix}\right|
\overset{(B)}{=}
\left|\begin{matrix}
1 &1 &1\\
0 &1 &-4\\
0 &3 &8
\end{matrix}\right|=20.
$$
c. General solution of homogeneous equation is $y^*:= C_1e^t +C_2e^{2t} +C_3e^{-3t}$. Special solution is
$\bar{y}:=Ate^{-3t}$ where $AL'(-3)=4$, $L'(k)=3k^2-7\implies L'(-3)=20\implies A=5$. So
$$
y= 5te^{-3t}+ C_1e^t +C_2e^{2t} +C_3e^{-3t}.
$$
