Let $\vec{a}=2 \hat{i}-\hat{j}+\hat{k}, \vec{b}=\hat{i}+2 \hat{j}+\hat{k}$ and $\vec{c}=\hat{i}+\hat

Question

Let $\vec{a}=2 \hat{i}-\hat{j}+\hat{k}, \vec{b}=\hat{i}+2 \hat{j}+\hat{k}$ and $\vec{c}=\hat{i}+\hat{j}-2 \hat{k}$ be three vectors. A vector in the plane of $\vec{b}$ and $\vec{c}$ whose projection on $\vec{a}$ is of magnitude $\sqrt{\frac{2}{3}}$ is\

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(a, c) : $\vec{a}=2 \hat{i}-\hat{j}+\hat{k}$\ $\vec{b}=\hat{i}+2 \hat{j}-\hat{k}$\ $\vec{c}=\hat{i}+\hat{j}-2 \hat{k}$\ Any vector $\vec{r}$ in the plane of $\vec{b}$ and $\vec{c}$ is  \ $=\hat{i}+2 \hat{j}-\hat{k}+t(\hat{i}+\hat{j}-2 \hat{k})$\ $=(1+t) \hat{i}+(2+t) \hat{j}-(1+2 t) \hat{k}$\ Projection of $\vec{r}$ on $a$ is $\sqrt{\frac{2}{3}}$\ $	herefore \quad \frac{\vec{r} \cdot \vec{a}}{|\vec{a}|}=\sqrt{\frac{2}{3}}$\ $\Rightarrow \quad\left|\frac{2(1+t)-(2+t)-(1+2 t)}{\sqrt{6}}ight|=\sqrt{\frac{2}{3}}$\ $\Rightarrow\left|-\frac{(1+t)}{\sqrt{2} \sqrt{3}}ight|=\frac{\sqrt{2}}{\sqrt{3}} \Rightarrow t=1$ or $-3 	herefore t=-3$ Put $t=1$ and -3 in equation (i) we have\ $\vec{r}=-2 \hat{i}-\hat{j}+5 \hat{k}$ or $\vec{r}=2 \hat{i}+3 \hat{j}-3 \hat{k}$\