Let $ \vec{a} = 2\hat{i} - \hat{j} - \hat{k}, \vec{b} = \hat{i} + 3\hat{j} - \hat{k} $ and $ \vec{c} = 2\hat{i} + \hat{j} + 3\hat{k} $. Let $ \vec{v} $ be the vector in the plane of the vectors $ \vec{a} $ and $ \vec{b} $, such that the length of its projection on the vector $ \vec{c} $ is $ \frac{1}{\sqrt{14}} $. Then $ |\vec{v}| $ is equal to

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

$ \vec{v} = (2x + y)\hat{i} + (3y - x)\hat{j} + (-x - y)\hat{k} $

$ \frac{|\vec{v}\cdot \vec{c}|}{|\vec{c}|} = \frac{1}{\sqrt{14}} $

$ \vec{v}\cdot \vec{c} = 2(2x + y) + (3y - x) + 3(-x - y) $

$ = 2y $

$ \frac{|2y|}{\sqrt{14}} = \frac{1}{\sqrt{14}} \Rightarrow |2y| = 1 $

$ |\vec{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.