Perpendicular from angle bisectors?

Inscribe circle $\Gamma$ inside triangle $ABC$ . Since $I$ is the intersection of two angle bisectors, it is the incenter of the triangle, and therefore is the centre of $\Gamma$ .

Also, note that since $IP\perp AC$ , $P$ must be the point where $\Gamma$ touches $AC$ . This is because $AC$ is tangent to the circle which means $IP$ is a radius of the circle. There is only one point $P$ on $AC$ where this can happen - the point of tangency.

Now, let the circle touch side $AC$ at $P$ , $AB$ at $Q$ , $BC$ at $R$ . Also, let $x=AP=AQ, y=BQ=BR, z=CP=CR$ . Using the side lengths of the triangle, we get $x+y=5$ , $y+z=6$ , $z+x=7$ . We can easily solve these equations to get $x=AP=\boxed{3}$ .