Lines \(\underline{r}=(1,3,5)+k(-1,2,3), k \in R\) and \(\underline{r}=(1,3,1)+\mathrm{k}(1,-2,3), \mathrm{k} \in \mathrm{R}\) are (A) coincident (B) parallel (C) skew (D) perpendicular

For the given lines: Line 1: \(\underline{r}=(1,3,5)+k(-1,2,3)\) Line 2: \(\underline{r}=(1,3,1)+\mathrm{k}(1,-2,3)\) The direction vectors are as follows: Direction vector of Line 1: \(\underline{v_1}=(-1,2,3)\) Direction vector of Line 2: \(\underline{v_2}=(1,-2,3)\)

Two vectors are parallel if one is a scalar multiple of the other. Let's check whether \(\underline{v_1}\) is a scalar multiple of \(\underline{v_2}\): \(\underline{v_1} = -1 \cdot \underline{v_2}\) \((-1,2,3) = -1 \cdot (1,-2,3)\) This condition holds, so the direction vectors of the lines are parallel.

Let's check if Line 1 and Line 2 share a common point. For this, we will equate the corresponding coordinates of their parametric equations and solve to find a common solution for k and \(\mathrm{k}\): For Line 1: \(x = 1 - k \) \(y = 3 + 2k\) \(z = 5 + 3k\) For Line 2: \(x = 1 + \mathrm{k}\) \(y = 3 - 2\mathrm{k}\) \(z = 1 + 3\mathrm{k}\) After equating the corresponding coordinates of the lines: \(1 - k = 1 + \mathrm{k}\) \(3 + 2k = 3 - 2\mathrm{k}\) \(5 + 3k = 1 + 3\mathrm{k}\) Now, by solving these equations, we get: From the first equation: \(k = - \mathrm{k}\) Substituting \(k\) as \(-\mathrm{k}\) in the second equation: \(3 + 2(-\mathrm{k}) = 3 - 2\mathrm{k}\) \(-2\mathrm{k} = -2\mathrm{k}\) This equation is always true. So, there is no unique solution for the third equation. Therefore, the lines share an infinite number of common points.

Since the direction vectors of the lines are parallel and they share an infinite number of common points, we can conclude that the lines are coincident. Thus, the correct option is: (A) coincident