Let $\vec{d}=\vec{b}+\lambda \vec{c}$
$$
\begin{aligned}
& \therefore \vec{d}=\hat{i}+2 \hat{j}-\hat{k}+\lambda(\hat{i}+\hat{j}-2 \hat{k}) \\
& =(1+\lambda) \hat{i}+(2+\lambda) \hat{j}-(1+2 \lambda) \hat{k}
\end{aligned}
$$
If $\theta$ be the angle between $\vec{d}$ and $\vec{a}$, then projection of $\vec{d}$ or $(\vec{b}+\lambda \vec{c})$ on $\vec{a}$
$$
\begin{aligned}
& =|\vec{d}| \cos \theta \\
& =|\vec{d}|\left(\frac{\vec{d} \cdot \vec{a}}{|\vec{d}||\vec{a}|}\right)=\frac{\vec{d} \cdot \vec{a}}{|\vec{a}|} \\
& =\frac{2(\lambda+1)-(\lambda+2)-(2 \lambda+1)}{\sqrt{4+1+1}} \\
& =\frac{-\lambda-1}{\sqrt{6}}
\end{aligned}
$$
But projection of $\vec{d}$ on $\vec{a}=\sqrt{\frac{2}{3}}$
$$
\therefore-\frac{\lambda+1}{\sqrt{6}}=\sqrt{\frac{2}{3}} \Rightarrow \frac{\lambda^2+2 \lambda+1}{6}=\frac{2}{3}
$$
$$
\begin{aligned}
& \Rightarrow \lambda^2+2 \lambda-3=0 \Rightarrow \lambda^2+3 \lambda-\lambda-3=0 \\
& \Rightarrow \lambda(\lambda+3)-1(\lambda+3)=0, \Rightarrow \lambda=1,-3
\end{aligned}
$$
when $\lambda=1$, then $\vec{b}+\lambda \vec{c}=2 \hat{i}+3 \hat{j}-3 \hat{k}$
when $\lambda=-3$, then $\vec{b}+\lambda \vec{c}=-2 \hat{i}-\hat{j}+5 \hat{k}$