On plane $\alpha$, there is a right isosceles triangle ABC with $\angle \mathrm { A } = 90 ^ { \circ }$ and $\overline { \mathrm { BC } } = 6$. A point P outside plane $\alpha$ is at a distance of 4 from the plane, and the foot of the perpendicular from P to plane $\alpha$ is point A. What is the distance from point P to line BC? [3 points]\\
(1) $3 \sqrt { 2 }$\\
(2) 5\\
(3) $3 \sqrt { 3 }$\\
(4) $4 \sqrt { 2 }$\\
(5) 6