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

\(
|\overline{\mathrm{u}}|=1,|\overline{\mathrm{v}}|=2,|\overline{\mathrm{w}}|=3
\)
According to the given condition,
(Projection of \(\bar{v}\) along \(\bar{u}\) )
\(
=(\text { Projection of } \bar{w} \text { along } \bar{u})
\)
\(
\begin{array}{ll}
\therefore & \frac{\overline{\mathrm{v}} \cdot \overline{\mathrm{u}}}{|\overline{\mathrm{u}}|}=\frac{\overline{\mathrm{w}} \cdot \overline{\mathrm{u}}}{|\overline{\mathrm{u}}|} \\
\therefore & \overline{\mathrm{v}} \cdot \overline{\mathrm{u}}=\overline{\mathrm{w}} \cdot \overline{\mathrm{u}} \\
\therefore & (\overline{\mathrm{w}}-\overline{\mathrm{v}}) \cdot \overline{\mathrm{u}}=0
\end{array}
\)
Now consider, \(|\overline{\mathrm{u}}-\overline{\mathrm{v}}+\overline{\mathrm{w}}|=\sqrt{|\overline{\mathrm{u}}+\overline{\mathrm{w}}-\overline{\mathrm{v}}|^2}\)
\(
\begin{aligned}
& =\sqrt{|\overline{\mathrm{u}}|^2+|\overline{\mathrm{w}}-\overline{\mathrm{v}}|^2+2 \overline{\mathrm{u}} \cdot(\overline{\mathrm{w}}-\overline{\mathrm{v}})} \\
& =\sqrt{(1)^2+|\overline{\mathrm{w}}-\overline{\mathrm{v}}|^2+0} \\
& =\sqrt{1+|\overline{\mathrm{w}}|^2+|\overline{\mathrm{v}}|^2-2(\overline{\mathrm{w}} \cdot \overline{\mathrm{v}})} \\
& =\sqrt{1+9+4+0}
\end{aligned}
\)
\(\ldots[\because \overline{\mathrm{w}}\) and \(\overline{\mathrm{v}}\) are perpendicular \(]\)
\(
=\sqrt{14}
\)