Discussion
A straight line passing through (6,1,3) meets the line \(\frac{x - 1}{2} = \frac{y}{1} = \frac{z - 2}{3}\) at Q. If the lines are perpendicular to each other, then the coordinates of \(Q\) are
Answer: C
\(Q = \) (3,1,5).
A general point of the given line is \(Q = (1 + 2t,\ t,\ 2 + 3t)\). With \(P = (6, 1, 3)\), \(\vec{PQ} = (2t - 5,\ t - 1,\ 3t - 1)\), which must be perpendicular to the direction (2, 1, 3):
\[ \begin{aligned} 2(2t - 5) + (t - 1) + 3(3t - 1) &= 0 \\ 14t - 14 &= 0 \Rightarrow t = 1 \end{aligned} \]
So \(Q = (3, 1, 5)\). The other points do not lie on the given line: for (2,1,3), \(\frac{2 - 1}{2} = \frac{1}{2}\) but \(\frac{y}{1} = 1\).
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