Which point of the line $\frac{x - 2}{2} = \frac{y + 3}{-2} = \frac{z - 1}{1}$ is at a distance of 6 from $(2, -3, 1)$ and has $z > 1$?

Explanation

1. 01

   A point of the line is $(2 + 2\lambda, -3 - 2\lambda, 1 + \lambda)$, at distance $\sqrt{4 + 4 + 1}\,\lvert \lambda \rvert = 3\lvert \lambda \rvert$ from $(2, -3, 1)$.

2. 02

   $3\lvert \lambda \rvert = 6$ gives $\lambda = \pm 2$, and $z > 1$ needs $\lambda = 2$: the point $(6, -7, 3)$.