Find the point of intersection of the lines f | vedclass

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

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

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(A) Let the first line be $\frac{x - 4}{5} = \frac{y - 1}{2} = \frac{z}{1} = s$. Then $x = 5s + 4, y = 2s + 1, z = s$. Let the second line be $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} = t$. Then $x = 2t + 1, y = 3t + 2, z = 4t + 3$. For intersection,$5s + 4 = 2t + 1 \implies 5s - 2t = -3$ (Eq. $1$). $2s + 1 = 3t + 2 \implies 2s - 3t = 1$ (Eq. $2$). Solving Eq. $1$ and Eq. $2$: Multiply Eq. $1$ by $3$ and Eq. $2$ by $2$: $15s - 6t = -9$ and $4s - 6t = 2$. Subtracting gives $11s = -11 \implies s = -1$. Substituting $s = -1$ into $x = 5s + 4, y = 2s + 1, z = s$,we get $x = 5(-1) + 4 = -1, y = 2(-1) + 1 = -1, z = -1$. Checking with the second line: $x = 2t + 1, y = 3t + 2, z = 4t + 3$. For $x = -1, 2t + 1 = -1 \implies t = -1$. Then $y = 3(-1) + 2 = -1$ and $z = 4(-1) + 3 = -1$. The point of intersection is $(-1, -1, -1)$.

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

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