The point of intersection of the lines frac{x | vedclass

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(A) Let the given lines be:$\frac{x - 5}{3} = \frac{y - 7}{-1} = \frac{z + 2}{1} = r_1$$\frac{x + 3}{-36} = \frac{y - 3}{2} = \frac{z - 6}{4} = r_2$Fr

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$(21, 5/3, 10/3)$

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(A) Let the given lines be:$\frac{x - 5}{3} = \frac{y - 7}{-1} = \frac{z + 2}{1} = r_1$$\frac{x + 3}{-36} = \frac{y - 3}{2} = \frac{z - 6}{4} = r_2$From the first line,any point is $(3r_1 + 5, -r_1 + 7, r_1 - 2)$.From the second line,any point is $(-36r_2 - 3, 2r_2 + 3, 4r_2 + 6)$.Equating the coordinates:$3r_1 + 5 = -36r_2 - 3 \implies 3r_1 + 36r_2 = -8$ (Eq. $1$)$-r_1 + 7 = 2r_2 + 3 \implies r_1 + 2r_2 = 4 \implies r_1 = 4 - 2r_2$ (Eq. $2$)Substitute $r_1$ in Eq. $1$:$3(4 - 2r_2) + 36r_2 = -8$$12 - 6r_2 + 36r_2 = -8$$30r_2 = -20 \implies r_2 = -2/3$Now,$r_1 = 4 - 2(-2/3) = 4 + 4/3 = 16/3$.Substituting $r_1$ into the first line coordinates:$x = 3(16/3) + 5 = 16 + 5 = 21$$y = -(16/3) + 7 = (-16 + 21)/3 = 5/3$$z = (16/3) - 2 = (16 - 6)/3 = 10/3$The point of intersection is $(21, 5/3, 10/3)$.

The point of intersection of the lines $\frac{x - 5}{3} = \frac{y - 7}{-1} = \frac{z + 2}{1}$ and $\frac{x + 3}{-36} = \frac{y - 3}{2} = \frac{z - 6}{4}$ is

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