The coordinates of the point,where the line t | vedclass

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

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$\left(\frac{17}{3}, 0, \frac{23}{3}\right)$

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(B) The equation of the line passing through $A(3, 4, 1)$ and $B(5, 1, 6)$ is given by $\frac{x-3}{5-3} = \frac{y-4}{1-4} = \frac{z-1}{6-1}$.This simplifies to $\frac{x-3}{2} = \frac{y-4}{-3} = \frac{z-1}{5} = k$.Any point on the line is $(2k+3, -3k+4, 5k+1)$.Since the line crosses the $XZ$-plane,the $y$-coordinate must be $0$.So,$-3k+4 = 0$,which gives $k = \frac{4}{3}$.Substituting $k = \frac{4}{3}$ into the coordinates:$x = 2(\frac{4}{3}) + 3 = \frac{8}{3} + 3 = \frac{17}{3}$.$z = 5(\frac{4}{3}) + 1 = \frac{20}{3} + 1 = \frac{23}{3}$.Thus,the required point is $\left(\frac{17}{3}, 0, \frac{23}{3}\right)$.

The coordinates of the point,where the line through $A(3, 4, 1)$ and $B(5, 1, 6)$ crosses the $XZ$-plane,are

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