To solve this problem, I need to find the point or points on the line $x= -3$ that are 10 units away from $(5,2)$. I can use the distance formula to write an equation for this condition: $\sqrt{(-3-5)^2+(y-2)^2}=10$. Squaring both sides, I get $64+(y-2)^2=100$. Subtracting 64 from both sides, I get $(y-2)^2=36$. Taking the square root of both sides, I get $y-2=\pm 6$. Adding 2 to both sides, I get $y=\pm 6+2$. So there are two possible values for $y$: 8 and -4. The product of these values is 8 times -4, which is -32. 
Point $P$ lies on the line $x= -3$ and is 10 units from the point $(5,2)$. Find the product of all possible $y$-coordinates that satisfy the given conditions.