Solution
Any vector $\vec{v}$ in plane of $\vec{a} \& \vec{b}$
\[
\begin{aligned}
& \vec{v}=x \vec{a}+y \vec{b}=x(2 \hat{i}-\hat{j}-\hat{k})+y(\hat{i}+3 \hat{j}-\hat{k}) \\
& \Rightarrow \vec{v}=(2 x+y) \hat{i}+(3 y-x) \hat{j}-(x+y) \hat{k}
\end{aligned}
\]
Projection length $\overrightarrow{\mathrm{v}}$ on $\overrightarrow{\mathrm{c}}=\frac{|\overrightarrow{\mathrm{v}} \cdot \overrightarrow{\mathrm{c}}|}{|\overrightarrow{\mathrm{c}}|}=\frac{1}{\sqrt{14}}$
\[
\begin{aligned}
& \Rightarrow \frac{|2(2 x+y)+1(3 y-x)-7(x+y)|}{\sqrt{4+1+9}}=\frac{1}{\sqrt{14}} \\
& \Rightarrow|4 x+2 y+3 y-x-3 x-3 y|=1 \\
& \Rightarrow|2 y|=1 \Rightarrow y= \pm \frac{1}{2} \\
& |\vec{V}|=\sqrt{(2 x+y)^{2}+(3 y-x)^{2}+(x+y)^{2}} \\
& =\sqrt{4 x^{2}+y^{2}+4 x y+9 y^{2}+x^{2}-6 x y+x^{2}+y^{2}+2 x y} \\
& =\sqrt{6 x^{2}+11 y^{2}}=\sqrt{6 x^{2}+\frac{11}{4}}=\frac{\sqrt{24 x^{2}+11}}{2}
\end{aligned}
\]
(1) $\frac{\sqrt{24 \mathrm{x}^{2}+11}}{2}=\frac{\sqrt{21}}{2} \Rightarrow 24 \mathrm{x}^{2}+11=21$
\[
\Rightarrow \mathrm{x}^{2}=\frac{10}{24} \Rightarrow \mathrm{x}= \pm \sqrt{\frac{5}{12}}
\]
(2) $\frac{\sqrt{24 \mathrm{x}^{2}+11}}{2}=13 \Rightarrow 24 \mathrm{x}^{2}+11=169 \times 4$
\[
\Rightarrow x \approx \pm 5.26
\]
(3) $\frac{\sqrt{24 \mathrm{x}^{2}+11}}{2}=\frac{\sqrt{35}}{2} \Rightarrow \mathrm{x}=1$
(4) $\frac{\sqrt{24 \mathrm{x}^{2}+11}}{2}=7 \Rightarrow 24 \mathrm{x}^{2}+11=4 \times 49$
\[
\Rightarrow \mathrm{x} \approx \pm 2.27 \text { Answer(1,2,3,4) }
\]
.
