Let $\overrightarrow{a}=\hat{i}-2 \hat{j}+3 \hat{k}$,$\overrightarrow{b}=2 \hat{i}+3 \hat{j}-\hat{k}$ and $\overrightarrow{c}=\lambda \hat{i}+\hat{j}+(2 \lambda-1) \hat{k}$. If $\overrightarrow{c}$ is parallel to the plane containing $\overrightarrow{a}$ and $\overrightarrow{b}$,then $\lambda$ is equal to

• A
  $0$
• B
  $1$
• C
  $-1$
• D
  $2$