Since $\vec{a}=\hat{i}-2 \hat{j}+3 \hat{k}, \vec{b}=2 \hat{i}+3 \hat{j}-\hat{k}$ and $\vec{c}=\lambda \hat{i}+\hat{j}+(2 \lambda-1 \hat{k})$ are coplanar therefore $[\vec{a} \vec{b} \vec{c}]=0$
$$
\begin{aligned}
& \text { i.e., }\left|\begin{array}{ccc}
1 & 2 & \lambda \\
-2 & 3 & 1 \\
3 & -1 & 2 \lambda-1
\end{array}\right|=0 \\
& \Rightarrow 1(6 \lambda-2)-2(-4 \lambda-1)+\lambda(-7)=0 \\
& \Rightarrow(6 \lambda-2)+8 \lambda+2+2+2 \lambda-9 \lambda=0 \\
& \Rightarrow 7 \lambda=0 \Rightarrow \lambda=0
\end{aligned}
$$