The point of intersection of lines frac{x - 4 | vedclass

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(A) Let the first line be $\frac{x - 4}{5} = \frac{y - 1}{2} = \frac{z}{1} = k_1$. Then $x = 5k_1 + 4, y = 2k_1 + 1, z = k_1$.Let the second line be $

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$(-1, -1, -1)$

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(A) Let the first line be $\frac{x - 4}{5} = \frac{y - 1}{2} = \frac{z}{1} = k_1$. Then $x = 5k_1 + 4, y = 2k_1 + 1, z = k_1$.Let the second line be $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} = k_2$. Then $x = 2k_2 + 1, y = 3k_2 + 2, z = 4k_2 + 3$.For the point of intersection,we equate the coordinates:$5k_1 + 4 = 2k_2 + 1 \implies 5k_1 - 2k_2 = -3$ (Equation $1$)$2k_1 + 1 = 3k_2 + 2 \implies 2k_1 - 3k_2 = 1$ (Equation $2$)Solving these equations: Multiply Equation $1$ by $3$ and Equation $2$ by $2$:$15k_1 - 6k_2 = -9$$4k_1 - 6k_2 = 2$Subtracting the equations: $11k_1 = -11 \implies k_1 = -1$.Substituting $k_1 = -1$ into the first line coordinates: $x = 5(-1) + 4 = -1, y = 2(-1) + 1 = -1, z = -1$.Thus,the point of intersection is $(-1, -1, -1)$.

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

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