Interpret the given data

We have $|\vec u|=1,\quad |\vec v|=2,\quad |\vec w|=3.$

Also,

• projection of $\vec v$ along $\vec u$ equals projection of $\vec w$ along $\vec u$,
• $\vec v \perp \vec w$.

Use equality of projections

Projection (scalar) of a vector $\vec a$ along $\vec u$ is $\frac{\vec a\cdot \vec u}{|\vec u|}.$

Since $|\vec u|=1$, equality of projections gives $\vec v\cdot \vec u=\vec w\cdot \vec u.$

Expand the required magnitude

We need $|\vec u-\vec v+\vec w|.$

First square it: $|\vec u-\vec v+\vec w|^2=(\vec u-\vec v+\vec w)\cdot(\vec u-\vec v+\vec w).$

Expanding, $=|\vec u|^2+|\vec v|^2+|\vec w|^2-2\vec u\cdot \vec v+2\vec u\cdot \vec w-2\vec v\cdot \vec w.$

Substitute the given conditions

Since $\vec v\perp \vec w$, $\vec v\cdot \vec w=0.$

Also from step 2, $\vec u\cdot \vec v=\vec u\cdot \vec w,$ so $-2\vec u\cdot \vec v+2\vec u\cdot \vec w=0.$

Therefore, $|\vec u-\vec v+\vec w|^2=|\vec u|^2+|\vec v|^2+|\vec w|^2.$

Now substitute magnitudes: $=1^2+2^2+3^2=1+4+9=14.$

Take square root

$|\vec u-\vec v+\vec w|=\sqrt{14}.$

Check options

The correct option is $\boxed{\sqrt{14}}$ which is Option C.