$\begin{aligned} & \text {} \frac{1}{6}\left|\begin{array}{ccc}1 & -\lambda & 1 \\ \lambda & -1 & -1 \\ 1 & 1 & \lambda\end{array}\right|=2 \\ & \Rightarrow\left|(-\lambda+1)+\lambda\left(\lambda^2+1\right)+1(\lambda+1)\right|=12 \\ & \Rightarrow 2+\lambda\left(\lambda^2+1\right)=12 \\ & \Rightarrow \lambda\left(\lambda^2+1\right)=10=2 \times 5 \\ & \therefore \quad \lambda=2 . \\ & \therefore|\lambda \hat{i}-3 \lambda \hat{j}+3 \hat{k}|=|2 \hat{i}-6 \hat{j}+3 \hat{k}| \\ & =\sqrt{2^2+(-6)^2+3^2}=\sqrt{49}=7 .\end{aligned}$