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Let's express the general points on both lines using parameters $\lambda$ and $\mu$. Line 1: $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4} = \lambda$ $x = 2\lambda + 1, y = 3\lambda + 2, z = 4\lambda + 3$ Line 2: $\frac{x-4}{5}=\frac{y-1}{2}=z = \mu$ $x = 5\mu + 4, y = 2\mu + 1, z = \mu$
For the lines to intersect, the coordinates must be equal at some point. $2\lambda + 1 = 5\mu + 4$ (1) $3\lambda + 2 = 2\mu + 1$ (2) $4\lambda + 3 = \mu$ (3)
Substitute (3) into (1) and (2): $2\lambda + 1 = 5(4\lambda + 3) + 4$ $2\lambda + 1 = 20\lambda + 15 + 4$ $18\lambda = -18$ $\lambda = -1$ $3\lambda + 2 = 2(4\lambda + 3) + 1$ $3\lambda + 2 = 8\lambda + 6 + 1$ $5\lambda = -5$ $\lambda = -1$ Now, find $\mu$ using equation (3): $\mu = 4\lambda + 3 = 4(-1) + 3 = -1$
Using $\lambda = -1$ in the first line's equations: $x = 2(-1) + 1 = -1$ $y = 3(-1) + 2 = -1$ $z = 4(-1) + 3 = -1$ So, the intersection point is $(-1, -1, -1)$.
The given point is $(-1, -5, -10)$. The intersection point is $(-1, -1, -1)$. Distance $d = \sqrt{(-1 - (-1))^2 + (-5 - (-1))^2 + (-10 - (-1))^2}$ $d = \sqrt{(0)^2 + (-4)^2 + (-9)^2}$ $d = \sqrt{0 + 16 + 81}$ $d = \sqrt{97}$
Final Answer: $\sqrt{97}$