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

$\sqrt{14}$

$|\vec u|=1;|\vec v|=2;|\vec w|=3$; $\vec v.\vec u=\vec w.\vec u$ and $\vec v.\vec w=0$

$|\vec u-\vec v+\vec w|^2=|\vec u|^2+|\vec v|^2+|\vec w|^2-2\vec u.\vec v-2\vec v.\vec w+2\vec u.\vec w=1+4+9$

$⇒|\vec u-\vec v+\vec w|=\sqrt{14}$