jmc

Let $z=a+bi,$ where $a$ and $b$ are real numbers. Then $$\begin{array}{rl}\frac{z_3-z_1}{z_2-z_1}\cdot \frac{z-z_2}{z-z_3}& =\frac{60+16i}{-44i}\cdot \frac{(a-18)+(b-39)i}{(a-78)+(b-99)i}\\ & =\frac{-4+15i}{11}\cdot \frac{[(a-18)+(b-39)i][(a-78)-(b-99)i]}{(a-78{)}^2+(b-99{)}^2}.\end{array}$$This expression is real if and only if the imaginary part is 0. In other words, $$(-4+15i)[(a-18)+(b-39)i][(a-78)-(b-99)i]$$has imaginary part 0. In turn this is equivalent to $$(-4)(-(a-18)(b-99)+(a-78)(b-39))+15((a-18)(a-78)+(b-39)(b-99))=0.$$This simplifies to $a^2-112a+b^2-122b+4929=0.$ Completing the square, we get $$(a-56{)}^2+(b-61{)}^2=1928,$$so $$(a-56{)}^2=1928-(b-61{)}^2.$$When $b$ is maximized, the right-hand side is 0, and $a=\boxed{56}.$