Let the line L intersect the lines $$x - 2 = -y = z - 1$$, $$2(x + 1) = 2(y - 1) = z + 1$$ and be parallel to the line $$\frac{x-2}{3} = \frac{y-1}{1} = \frac{z-2}{2}$$. Then which of the following points lies on L?

General Points

• Point $$P$$ on $$L_1$$: $$( \lambda+2, -\lambda, \lambda+1 )$$

• Point $$Q$$ on $$L_2$$: $$( \frac{\mu}{2}-1, \frac{\mu}{2}+1, \mu-1 )$$

• Direction vector $$\vec{PQ}$$: $$( \frac{\mu}{2}-\lambda-3, \frac{\mu}{2}+\lambda+1, \mu-\lambda-2 )$$

Since $$\vec{PQ} \parallel (3, 1, 2)$$, the ratios must be equal:

$$\frac{\frac{\mu}{2}-\lambda-3}{3} = \frac{\frac{\mu}{2}+\lambda+1}{1} = \frac{\mu-\lambda-2}{2}$$

Solving these equations gives:

• $$\lambda = -4/3$$

• $$\mu = -2/3$$

Equation of Line $$L$$

Using $$\lambda = -4/3$$, point $$P$$ is $$(\frac{2}{3}, \frac{4}{3}, -\frac{1}{3})$$. The line equation is:

$$\frac{x - 2/3}{3} = \frac{y - 4/3}{1} = \frac{z + 1/3}{2} = k$$

Testing Option A $$(-\frac{1}{3}, 1, -1)$$:

• x: $$(-1/3 - 2/3) / 3 = -1/3$$

• y: $$(1 - 4/3) / 1 = -1/3$$

• z: $$(-1 + 1/3) / 2 = -1/3$$

All coordinates yield $$k = -1/3$$, so the point lies on $$L$$.

Correct Option: A