The line joining the points (3, 5, -7) and (- | vedclass

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(A) The equation of the line passing through points $(x_1, y_1, z_1) = (3, 5, -7)$ and $(x_2, y_2, z_2) = (-2, 1, 8)$ is given by: $\frac{x - 3}{-2 -

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$\left( 0, \frac{13}{5}, 2 \right)$

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(A) The equation of the line passing through points $(x_1, y_1, z_1) = (3, 5, -7)$ and $(x_2, y_2, z_2) = (-2, 1, 8)$ is given by: $\frac{x - 3}{-2 - 3} = \frac{y - 5}{1 - 5} = \frac{z - (-7)}{8 - (-7)}$ $\frac{x - 3}{-5} = \frac{y - 5}{-4} = \frac{z + 7}{15} = K$ Any point on this line is given by $(x, y, z) = (-5K + 3, -4K + 5, 15K - 7)$. Since the line meets the $yz$-plane,the $x$-coordinate must be $0$. $-5K + 3 = 0 \Rightarrow K = \frac{3}{5}$. Substituting $K = \frac{3}{5}$ into the coordinates: $x = -5(\frac{3}{5}) + 3 = 0$ $y = -4(\frac{3}{5}) + 5 = -\frac{12}{5} + \frac{25}{5} = \frac{13}{5}$ $z = 15(\frac{3}{5}) - 7 = 9 - 7 = 2$ Thus,the point of intersection is $\left( 0, \frac{13}{5}, 2 \right)$.

The line joining the points $(3, 5, -7)$ and $(-2, 1, 8)$ meets the $yz$-plane at point

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