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(C) Let the first line be $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4} = k_1$. Any point on this line is $(2k_1+1, 3k_1+2, 4k_1+3)$.Let the second line

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$\frac{x+1}{3}=\frac{y+1}{2}=\frac{z+1}{-1}$

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(C) Let the first line be $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4} = k_1$. Any point on this line is $(2k_1+1, 3k_1+2, 4k_1+3)$.Let the second line be $\frac{x-4}{5}=\frac{y-1}{2}=\frac{z}{1} = k_2$. Any point on this line is $(5k_2+4, 2k_2+1, k_2)$.Equating the coordinates for intersection: $2k_1+1 = 5k_2+4 \implies 2k_1 - 5k_2 = 3$ and $3k_1+2 = 2k_2+1 \implies 3k_1 - 2k_2 = -1$.Solving these,we get $k_1 = -1$ and $k_2 = -1$. The intersection point is $(-1, -1, -1)$.The line passes through $(-1, -1, -1)$ and $(2, 1, -2)$.The direction vector is $(2 - (-1), 1 - (-1), -2 - (-1)) = (3, 2, -1)$.The equation of the line is $\frac{x+1}{3} = \frac{y+1}{2} = \frac{z+1}{-1}$.

The equation of the line passing through the point of intersection of $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x-4}{5}=\frac{y-1}{2}=z$ and also through the point $(2,1,-2)$ is

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