Given that, $\overrightarrow{\mathbf{a}}=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$
$\overrightarrow{\mathbf{b}}=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-\hat{\mathbf{k}}$
and $\quad \overrightarrow{\mathbf{c}}=\lambda \hat{\mathbf{i}}+\hat{\mathbf{j}}+(2 \lambda-1) \hat{\mathbf{k}}$


Also, since $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$ lies in the same plane, then $(\vec{a} \times \vec{b})$ is perpendicular vector to this plane. Given that vector $\overrightarrow{\mathbf{c}}$ is parallel to the plane containing $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$, so vector $(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}})$ also perpendicular to the vector $\overrightarrow{\mathbf{c}}$ ie, $\left(\theta=90^{\circ}\right)$ So, $(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}) \cdot \overrightarrow{\mathbf{c}}$ should be equal to zero
or $(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}) \cdot \overrightarrow{\mathbf{c}}=0$ ...(i)
$\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}=\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & -2 & 3 \\ 2 & 3 & -1\end{array}\right|$
$=(2-9) \hat{\mathbf{i}}+(6+1) \hat{\mathbf{j}}+(3+4) \hat{\mathbf{k}}$
$=-7 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}+7 \hat{\mathbf{k}}$
Then from Eq. (i)
$(-7 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}+7 \hat{\mathbf{k}}) \cdot(\lambda \hat{\mathbf{i}}+\hat{\mathbf{j}}+(2 \lambda-1) \hat{\mathbf{k}})=0$
$\Rightarrow \quad-7 \lambda+7+7(2 \lambda-1)=0$
$\Rightarrow \quad-7 \lambda+7+14 \lambda-7=0$
$\Rightarrow \quad 7 \lambda=0$
$\Rightarrow \quad \lambda=0$
Hence, the value of $\lambda$ is 0 .