The coordinates of the point where the line p | vedclass

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(C) The equation of the line passing through $P(3, 4, 1)$ and $Q(5, 1, 6)$ is given by $\frac{x-3}{5-3} = \frac{y-4}{1-4} = \frac{z-1}{6-1} = k$. This

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$(\frac{13}{5}, \frac{23}{5}, 0)$

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(C) The equation of the line passing through $P(3, 4, 1)$ and $Q(5, 1, 6)$ is given by $\frac{x-3}{5-3} = \frac{y-4}{1-4} = \frac{z-1}{6-1} = k$. This simplifies to $\frac{x-3}{2} = \frac{y-4}{-3} = \frac{z-1}{5} = k$. Any point on this line is $(2k+3, -3k+4, 5k+1)$. Since the point lies on the $xy$-plane,its $z$-coordinate must be $0$. Therefore,$5k+1 = 0$,which gives $k = -\frac{1}{5}$. Substituting $k = -\frac{1}{5}$ into the coordinates: $x = 2(-\frac{1}{5}) + 3 = -\frac{2}{5} + \frac{15}{5} = \frac{13}{5}$. $y = -3(-\frac{1}{5}) + 4 = \frac{3}{5} + \frac{20}{5} = \frac{23}{5}$. $z = 0$. Thus,the point is $(\frac{13}{5}, \frac{23}{5}, 0)$.

The coordinates of the point where the line passing through $P(3, 4, 1)$ and $Q(5, 1, 6)$ crosses the $xy$-plane are:

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