The distance of line 3y−2z−1=0=3x−z+4 from the point (2,−1,6) is
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Official solution
Parametrize by z=t: point (3t−4,31+2t,t), direction (1,2,3) (after scaling by 3). At t=0: A=(−34,31,0).
\[\overrightarrow{AP}=\left(\tfrac{10}{3},-\tfrac43,6\right),\qquad \overrightarrow{AP}\cdot(1,2,3)=\tfrac{10}{3}-\tfrac83+18=\tfrac{56}{3}.\]
\[|\overrightarrow{AP}|^2=\tfrac{100}9+\tfrac{16}9+36=\tfrac{440}9,\qquad \frac{(\overrightarrow{AP}\cdot\vec d)^2}{|\vec d|^2}=\frac{(56/3)^2}{14}=\tfrac{224}9.\]
\[\text{Distance}^2=\tfrac{440}9-\tfrac{224}9=\tfrac{216}9=24 \implies \text{Distance}=\sqrt{24}=2\sqrt{6}.\]