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Let $M$ be the midpoint of the line segment joining $A(2,3,4)$ and $B(4,5,8)$. The coordinates of $M$ are given by: $$M = \left(\frac{2+4}{2}, \frac{3+5}{2}, \frac{4+8}{2}\right) = (3, 4, 6)$$
The line is perpendicular to the lines with direction ratios $(3, -16, 7)$ and $(3, 8, -5)$. Let the direction ratios of the required line be $(a, b, c)$. Since the line is perpendicular to both given lines, we have: $$3a - 16b + 7c = 0$$ $$3a + 8b - 5c = 0$$ Subtracting the second equation from the first, we get: $$-24b + 12c = 0 \implies 2b = c$$ Substituting $c = 2b$ in the second equation: $$3a + 8b - 5(2b) = 0 \implies 3a + 8b - 10b = 0 \implies 3a = 2b \implies a = \frac{2}{3}b$$ Let $b = 3k$, then $a = 2k$ and $c = 6k$. Thus, the direction ratios of the required line are $(2k, 3k, 6k)$, or simply $(2, 3, 6)$.
The equation of the line passing through the point $M(3, 4, 6)$ and having direction ratios $(2, 3, 6)$ is given by: $$\frac{x - 3}{2} = \frac{y - 4}{3} = \frac{z - 6}{6}$$
Final Answer: $\frac{x - 3}{2} = \frac{y - 4}{3} = \frac{z - 6}{6}$
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