The point where the line through $A = (3, - 2, 7)$ and $B = (13, 3, - 8)$ meets the xy-plane

(\frac{23}{3}, \frac{1}{3}, 0)

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(\frac{23}{6}, \frac{1}{6}, 0)

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(\frac{23}{3}, \frac{1}{3}, 1)

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(\frac{23}{3}, \frac{1}{3}, 3)

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Solution

The correct option is A $(\frac{23}{3}, \frac{1}{3}, 0)$
Equation of line through $A (3, - 2, 7)$ and $B (13, 3, - 8)$ is:
$\frac{x - 3}{10} = \frac{y + 2}{5} = \frac{z - 7}{- 15}$
When the line meets $x - y$ plane $\Rightarrow z = 0$
$∴ \frac{x - 3}{10} = \frac{y + 2}{5} = \frac{7}{15} \Rightarrow x = \frac{23}{3}, y = \frac{1}{3} \Rightarrow (x, y, z) = (\frac{23}{3}, \frac{1}{3}, 0)$

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