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

Given vectors u, v, w with magnitudes $|\overline{\mathrm{u}}|=1$, $|\overline{\mathrm{v}}|=2$, $|\overline{\mathrm{w}}|=3$, and conditions that the projection of v and w along u are equal, and that v and w are perpendicular.

The projection equality implies $\frac{\overline{\mathrm{v}} \cdot \overline{\mathrm{u}}}{|\overline{\mathrm{u}}|} = \frac{\overline{\mathrm{w}} \cdot \overline{\mathrm{u}}}{|\overline{\mathrm{u}}|}$, so $\overline{\mathrm{v}} \cdot \overline{\mathrm{u}} = \overline{\mathrm{w}} \cdot \overline{\mathrm{u}}$.
Since $\overline{\mathrm{v}} \cdot \overline{\mathrm{w}} = 0$, we compute the squared magnitude:

$|\overline{\mathrm{u}}-\overline{\mathrm{v}}+\overline{\mathrm{w}}|^2 = (\overline{\mathrm{u}}-\overline{\mathrm{v}}+\overline{\mathrm{w}}) \cdot (\overline{\mathrm{u}}-\overline{\mathrm{v}}+\overline{\mathrm{w}})$
$= |\overline{\mathrm{u}}|^2 + |\overline{\mathrm{v}}|^2 + |\overline{\mathrm{w}}|^2 - 2(\overline{\mathrm{u}} \cdot \overline{\mathrm{v}}) + 2(\overline{\mathrm{u}} \cdot \overline{\mathrm{w}}) - 2(\overline{\mathrm{v}} \cdot \overline{\mathrm{w}})$
$= 1 + 4 + 9 - 2(\overline{\mathrm{u}} \cdot \overline{\mathrm{v}}) + 2(\overline{\mathrm{u}} \cdot \overline{\mathrm{v}}) - 0 = 14$

Thus $|\overline{\mathrm{u}}-\overline{\mathrm{v}}+\overline{\mathrm{w}}| = \sqrt{14}$, matching option A.