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

Since the projection of $$\vec v$$ along $$\vec u$$ is equal to that of $$\vec w$$ along $$\vec u$$,

$$\frac{\vec v\cdot\vec u}{|\vec u|}=\frac{\vec w\cdot\vec u}{|\vec u|}$$

Since

$$|\vec u|=1$$

we get

$$\vec u\cdot\vec v=\vec u\cdot\vec w$$

Therefore,

$$\vec u\cdot(\vec w-\vec v)=0$$

Also,

$$\vec v\cdot\vec w=0$$

Now,

$$|\vec u-\vec v+\vec w|^2=(\vec u-\vec v+\vec w)\cdot(\vec u-\vec v+\vec w)$$

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

Using

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

and

$$\vec u\cdot\vec v=\vec u\cdot\vec w,\qquad \vec v\cdot\vec w=0$$

we obtain

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

$$=14$$

Hence,

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

Therefore,

$$\boxed{\sqrt{14}}$$