Let \(\overline{\mathrm{u}}, \overline{\mathrm{v}}\) and \(\overline{\mathrm{w}}\) be the vectors such that \(|\overline{\mathrm{u}}|=1\); \(|\overline{\mathrm{v}}|=2 ;|\overline{\mathrm{w}}|=3\). If the projection of \(\overline{\mathrm{v}}\) along \(\overline{\mathrm{u}}\) is equal to that of \(\bar{w}\) along \(\bar{u}\) and \(\bar{v}, \bar{w}\) are perpendicular to each other, then \(|\bar{u}-\bar{v}+\bar{w}|\) is equal to

Projection of \(\overline{\mathrm{v}}\) along \(\overline{\mathrm{u}}=\) Projection of \(\overline{\mathrm{w}}\) along \(\overline{\mathrm{u}}\)
\(\begin{aligned}
& \Rightarrow \frac{\overline{\mathrm{v}} \cdot \overline{\mathrm{u}}}{|\overline{\mathrm{u}}|}=\frac{\overline{\mathrm{w}} \cdot \overline{\mathrm{u}}}{|\overline{\mathrm{u}}|} \\
& \Rightarrow \overline{\mathrm{v}} \cdot \overline{\mathrm{u}}=\overline{\mathrm{w}} \cdot \overline{\mathrm{u}}... (i)
\end{aligned}\)
Also, \(\bar{v}\) and \(\bar{w}\) are perpendicular to each other.
\(\therefore \quad \overline{\mathrm{v}} \cdot \overline{\mathrm{w}}=0... (ii)\)
Now, \(|\overline{\mathrm{u}}-\overline{\mathrm{v}}+\overline{\mathrm{w}}|^2=|\overline{\mathrm{u}}|^2+|\overline{\mathrm{v}}|^2+|\overline{\mathrm{w}}|^2-2(\overline{\mathrm{u}} \cdot \overline{\mathrm{v}})\)
\(-2(\overline{\mathrm{v}} \cdot \overline{\mathrm{w}})+2(\overline{\mathrm{u}} \cdot \overline{\mathrm{w}})\)
\(\Rightarrow|\overline{\mathrm{u}}-\overline{\mathrm{v}}+\overline{\mathrm{w}}|^2=1+4+9\)
\(\Rightarrow|\bar{u}-\bar{v}+\bar{w}|=\sqrt{14}\)