Let L be the line (x-1)/2=(y-2)/(-1)=(z+1)/3 . Match each line M in L…

The correct answer is Option (3) → (A)-(III), (B)-(IV), (C)-(II), (D)-(I)

L passes through (1, 2, −1) with direction ratios 2, −1, 3. A general point of L is \((1+2s,\ 2-s,\ -1+3s)\).

If the direction ratios of M are proportional to 2, −1, 3, the lines are parallel: the same line if a point of M is on L, otherwise distinct. If not, they either intersect or are skew.

(A) \(\displaystyle\frac{x-3}{-4}=\frac{y-1}{2}=\frac{z-2}{-6}\)

This matches (III) the same line. −4, 2, −6 is −2 times 2, −1, 3, and (3, 1, 2) is on L:

    \( \displaystyle \frac{3-1}{2} = 1,\quad \frac{1-2}{-1} = 1,\quad \frac{2+1}{3} = 1 \)

(B) \(\displaystyle\frac{x-2}{4}=\frac{y+1}{-2}=\frac{z}{6}\)

This matches (IV) parallel and distinct lines. 4, −2, 6 is 2 times 2, −1, 3, but (2, −1, 0) is not on L:

    \( \displaystyle \frac{2-1}{2} = \frac{1}{2},\quad \frac{-1-2}{-1} = 3 \)

(C) \(\displaystyle\frac{x-4}{1}=\frac{y-3}{2}=\frac{z-1}{-1}\)

This matches (II) intersecting lines. A point of M is \((4+t,\ 3+2t,\ 1-t)\):

    \(1+2s = 4+t\) and \(2-s = 3+2t\) give \(s = 1,\ t = -1\)

    z-coordinates: \(-1+3 = 2\) and \(1+1 = 2\), equal

So the lines meet at (3, 1, 2).

(D) \(\displaystyle\frac{x-2}{1}=\frac{y-1}{-1}=\frac{z}{2}\)

This matches (I) skew lines. 1, −1, 2 is not proportional to 2, −1, 3. A point of M is \((2+t,\ 1-t,\ 2t)\):

    \(1+2s = 2+t\) and \(2-s = 1-t\) give \(s = 0,\ t = -1\)

    z-coordinates: \(-1+0 = -1\) and \(2(-1) = -2\), not equal

So the lines neither meet nor are parallel.