Let overline{a}=hat{i}+hat{j}+hat{k},overline | vedclass

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(C) Given $\overline{a}=\hat{i}+\hat{j}+\hat{k}$,$\overline{b}=\hat{i}-\hat{j}+\hat{k}$ and $\overline{c}=\hat{i}-\hat{j}-\hat{k}$.Since $\overline{v}

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$3 \hat{i}-\hat{j}+3 \hat{k}$

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(C) Given $\overline{a}=\hat{i}+\hat{j}+\hat{k}$,$\overline{b}=\hat{i}-\hat{j}+\hat{k}$ and $\overline{c}=\hat{i}-\hat{j}-\hat{k}$.Since $\overline{v}$ lies in the plane of $\overline{a}$ and $\overline{b}$,we can write $\overline{v} = m\overline{a} + n\overline{b}$.$\overline{v} = m(\hat{i}+\hat{j}+\hat{k}) + n(\hat{i}-\hat{j}+\hat{k}) = (m+n)\hat{i} + (m-n)\hat{j} + (m+n)\hat{k} \quad \dots(i)$The projection of $\overline{v}$ on $\overline{c}$ is given by $\frac{\overline{v} \cdot \overline{c}}{|\overline{c}|} = \frac{1}{\sqrt{3}}$.$|\overline{c}| = \sqrt{1^2 + (-1)^2 + (-1)^2} = \sqrt{3}$.So,$\frac{(m+n)(1) + (m-n)(-1) + (m+n)(-1)}{\sqrt{3}} = \frac{1}{\sqrt{3}}$.$m+n - m+n - m-n = 1 \implies n-m = 1 \implies n = m+1$.Substituting $n = m+1$ into $(i)$:$\overline{v} = (m+m+1)\hat{i} + (m-(m+1))\hat{j} + (m+m+1)\hat{k} = (2m+1)\hat{i} - \hat{j} + (2m+1)\hat{k}$.For $m=1$,$\overline{v} = 3\hat{i} - \hat{j} + 3\hat{k}$,which matches option $C$.

Let $\overline{a}=\hat{i}+\hat{j}+\hat{k}$,$\overline{b}=\hat{i}-\hat{j}+\hat{k}$ and $\overline{c}=\hat{i}-\hat{j}-\hat{k}$ be three vectors. $A$ vector $\overline{v}$ in the plane of $\overline{a}$ and $\overline{b}$,whose projection on $\overline{c}$ is $\frac{1}{\sqrt{3}}$,is given by

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