Question

$\overrightarrow{v} = x\overrightarrow{a} + y\overrightarrow{b} = x(2\widehat{i} - \widehat{j} - \widehat{k}) + y(\widehat{i} + 3\widehat{j} - \widehat{k})$

$${\overrightarrow{v} = (2x + y)\widehat{i} + (3y - x)\widehat{j} + ( - x - y)\widehat{k} }{\left| \frac{\overrightarrow{v} \cdot \overrightarrow{C}}{|\overrightarrow{c}|} \right| = \frac{1}{\sqrt{14}} }{\overrightarrow{v} \cdot \overrightarrow{c} = 2(2x + y) + 3y - x - 3x - 3y }{= 2y }{\left| \frac{2y}{\sqrt{14}} \right| = \frac{1}{\sqrt{14}} \Rightarrow |2y| = 1 }{|\overrightarrow{v}| = \sqrt{(2x + y)^{2} + (3y - x)^{2} + (x + y)^{2}} }{= \sqrt{6x^{2} + 11y^{2} + 4xy - 6xy + 2xy} }{= \sqrt{6x^{2} + \frac{11}{4}} = \frac{\sqrt{24x^{2} + 11}}{2} }$$Now if we take $x^{2} = 1$ then option $\frac{\sqrt{35}}{2}$ matches most probably NTA thought could been this. Ans (Bonus), NTA (3)