Question:

Equation of line through point $(-1, 2, 1)$ parallel to $\frac{x-1}{2} = \frac{y+1}{3} = \frac{z}{4}$ is: $$\frac{x+1}{2} = \frac{y-2}{3} = \frac{z-1}{4} = \lambda$$ So, $\begin{cases} x = 2\lambda - 1 \\ y = 3\lambda + 2 \\ z = 4\lambda + 1 \end{cases}$ For the line $\frac{x+2}{3} = \frac{y-3}{2} = \frac{z-4}{1} = \mu$: $\begin{cases} x = 3\mu - 2 \\ y = 2\mu + 3 \\ z = \mu + 4 \end{cases}$ For intersection point $P$: $$2\lambda - 1 = 3\mu - 2$$ $$3\lambda + 2 = 2\mu + 3$$ $$4\lambda + 1 = \mu + 4$$ From the third equation: $\mu = 4\lambda - 3$ Substituting into the second equation: $$3\lambda + 2 = 2(4\lambda - 3) + 3$$ $$3\lambda + 2 = 8\lambda - 6 + 3$$ $$3\lambda + 2 = 8\lambda - 3$$ $$5 = 5\lambda$$ $$\lambda = 1$$ Therefore $\mu = 1$. Point $P: x = 2(1) - 1 = 1$, $y = 3(1) + 2 = 5$, $z = 4(1) + 1 = 5$ So $P(1, 5, 5)$ and $Q(4, -5, 1)$ $$PQ = \sqrt{(4-1)^2 + (-5-5)^2 + (1-5)^2} = \sqrt{9 + 100 + 16} = \sqrt{125} = 5\sqrt{5}$$