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