Two lines frac{x - 3}{1} = frac{y + 1}{3} = f | vedclass

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(A) Let the points on the two lines be $P_1 = (\lambda + 3, 3\lambda - 1, -\lambda + 6)$ and $P_2 = (7\alpha - 5, -6\alpha + 2, 4\alpha + 3)$.For the

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$(2, -4, -7)$

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(A) Let the points on the two lines be $P_1 = (\lambda + 3, 3\lambda - 1, -\lambda + 6)$ and $P_2 = (7\alpha - 5, -6\alpha + 2, 4\alpha + 3)$.For the lines to intersect at point $R$,we equate the coordinates:$\lambda + 3 = 7\alpha - 5 \Rightarrow \lambda - 7\alpha = -8$ (Equation $1$)$3\lambda - 1 = -6\alpha + 2 \Rightarrow 3\lambda + 6\alpha = 3 \Rightarrow \lambda + 2\alpha = 1$ (Equation $2$)Subtracting Equation $1$ from Equation $2$ gives $9\alpha = 9$,so $\alpha = 1$.Substituting $\alpha = 1$ into Equation $2$,we get $\lambda + 2(1) = 1$,so $\lambda = -1$.Substituting $\lambda = -1$ into the first line equation,we get $R = (-1 + 3, 3(-1) - 1, -(-1) + 6) = (2, -4, 7)$.The reflection of a point $(x, y, z)$ in the $xy$-plane is $(x, y, -z)$.Therefore,the reflection of $R(2, -4, 7)$ in the $xy$-plane is $(2, -4, -7)$.

Two lines $\frac{x - 3}{1} = \frac{y + 1}{3} = \frac{z - 6}{-1}$ and $\frac{x + 5}{7} = \frac{y - 2}{-6} = \frac{z - 3}{4}$ intersect at the point $R$. The reflection of $R$ in the $xy$-plane has coordinates

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