The lines frac{x - 1}{2} = frac{y + 1}{2} = f | vedclass

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(B) Let the general point on the first line $L_1$ be $P(r)$.$\frac{x - 1}{2} = \frac{y + 1}{2} = \frac{z - 1}{4} = r$$x = 2r + 1, y = 2r - 1, z = 4r +

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$(-2, -4, -5)$

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(B) Let the general point on the first line $L_1$ be $P(r)$.$\frac{x - 1}{2} = \frac{y + 1}{2} = \frac{z - 1}{4} = r$$x = 2r + 1, y = 2r - 1, z = 4r + 1$So,$P = (2r + 1, 2r - 1, 4r + 1)$.Since the lines intersect,this point must satisfy the equation of the second line $L_2$: $\frac{x - 3}{1} = \frac{y - 6}{2} = \frac{z}{1} = k$.Substituting the coordinates of $P$ into $L_2$:$\frac{2r + 1 - 3}{1} = \frac{2r - 1 - 6}{2} = \frac{4r + 1}{1}$$\frac{2r - 2}{1} = \frac{2r - 7}{2} = 4r + 1$Taking the first and third parts: $2r - 2 = 4r + 1 \Rightarrow -3 = 2r \Rightarrow r = -\frac{3}{2}$.Now,substitute $r = -\frac{3}{2}$ into the coordinates of $P$:$x = 2(-\frac{3}{2}) + 1 = -3 + 1 = -2$$y = 2(-\frac{3}{2}) - 1 = -3 - 1 = -4$$z = 4(-\frac{3}{2}) + 1 = -6 + 1 = -5$Thus,the intersection point is $(-2, -4, -5)$.

The lines $\frac{x - 1}{2} = \frac{y + 1}{2} = \frac{z - 1}{4}$ and $\frac{x - 3}{1} = \frac{y - 6}{2} = \frac{z}{1}$ intersect each other at point

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