b
Unit vector $\hat{\mathrm{u}}=\mathrm{x} \hat{\mathrm{i}}+\mathrm{y} \hat{\mathrm{j}}+\mathrm{z} \hat{\mathrm{k}}$
$ \overrightarrow{\mathrm{p}}_1=\frac{1}{\sqrt{2}} \hat{\mathrm{i}}+\frac{1}{\sqrt{2}} \hat{\mathrm{k}}, \overrightarrow{\mathrm{p}}_2=\frac{1}{\sqrt{2}} \hat{\mathrm{j}}+\frac{1}{\sqrt{2}} \hat{\mathrm{k}}$
$ \overrightarrow{\mathrm{p}}_3=\frac{1}{\sqrt{2}} \hat{\mathrm{i}}+\frac{1}{\sqrt{2}} \hat{\mathrm{j}}$
Now angle between $\hat{\mathrm{u}}$ and $\overrightarrow{\mathrm{p}}_1=\frac{\pi}{2}$ $\hat{\mathrm{u}} \cdot \overrightarrow{\mathrm{p}}_1=0 \Rightarrow \frac{\mathrm{x}}{\sqrt{2}}+\frac{\mathrm{z}}{\sqrt{2}}=0$
$\Rightarrow \mathrm{x}+\mathrm{z}=0$ $.........(i)$
Angle between $\hat{\mathrm{u}}$ and $\overrightarrow{\mathrm{p}}_2=\frac{\pi}{3}$
$\hat{\mathrm{u}} \cdot \overrightarrow{\mathrm{p}}_2=|\hat{\mathrm{u}}| \cdot\left|\overrightarrow{\mathrm{p}}_2\right| \cos \frac{\pi}{3}$
$ \Rightarrow \frac{\mathrm{y}}{\sqrt{2}}+\frac{\mathrm{z}}{\sqrt{2}}=\frac{1}{2} \Rightarrow \mathrm{y}+\mathrm{z}=\frac{1}{\sqrt{2}}$ $..............(ii)$
Angle between $\hat{\mathrm{u}}$ and $\overrightarrow{\mathrm{p}}_3=\frac{2 \pi}{3}$
$ \hat{\mathrm{u}} \cdot \overrightarrow{\mathrm{p}}_3=|\hat{\mathrm{u}}| \cdot\left|\overrightarrow{\mathrm{p}}_3\right| \cos \frac{2 \pi}{3} $
$ \Rightarrow \frac{\mathrm{x}}{\sqrt{2}}+\frac{4}{\sqrt{2}}=\frac{-1}{2} \Rightarrow \mathrm{x}+\mathrm{y}=\frac{-1}{\sqrt{2}}$ $............(iii)$
from equation $(i)$, $(ii)$ and $(iii)$ we get
$x=\frac{-1}{\sqrt{2}} \quad y=0 \quad z=\frac{1}{\sqrt{2}}$
Thus $\hat{\mathrm{u}}-\overrightarrow{\mathrm{v}}=\frac{-1}{\sqrt{2}} \hat{\mathrm{i}}+\frac{1}{\sqrt{2}} \hat{\mathrm{k}}-\frac{1}{\sqrt{2}} \hat{\mathrm{i}}-\frac{1}{\sqrt{2}} \hat{\mathrm{j}}-\frac{1}{\sqrt{2}} \hat{\mathrm{k}}$
$\hat{\mathrm{u}}-\overrightarrow{\mathrm{v}}=\frac{-2}{\sqrt{2}} \hat{\mathrm{i}}-\frac{1}{\sqrt{2}} \hat{\mathrm{j}} $
$ \therefore|\hat{\mathrm{u}}-\overrightarrow{\mathrm{v}}|^2=\left(\sqrt{\frac{4}{2}+\frac{1}{2}}\right)^2=\frac{5}{2}$