The lines $$\vec{r} = (\hat{i} - \hat{j}) + l(2\hat{i} + \hat{k})$$ and $$\vec{r} = (2\hat{i} - \hat{j}) + m(\hat{i} + \hat{j} - \hat{k})$$

We have two lines in vector form:

$$\vec{r} = (\hat{i} - \hat{j}) + l\,(2\hat{i} + \hat{k})$$

and

$$\vec{r} = (2\hat{i} - \hat{j}) + m\,(\hat{i} + \hat{j} - \hat{k}).$$

To decide whether the lines intersect, we assume that for some real numbers $$l$$ and $$m$$ the position vectors of a common point are equal. Hence we equate the corresponding Cartesian components.

First, we rewrite each vector equation in component form. Remember that $$\hat{i},\ \hat{j},\ \hat{k}$$ are unit vectors along the $$x, y, z$$ axes respectively.

For the first line:

$$\vec{r}_1 = (1)\hat{i} + (-1)\hat{j} + (0)\hat{k} + l\bigl(2\hat{i} + 0\hat{j} + 1\hat{k}\bigr).$$

So the coordinates of a general point on line 1 are

$$x_1 = 1 + 2l,\qquad y_1 = -1 + 0\cdot l = -1,\qquad z_1 = 0 + 1\cdot l = l.$$

For the second line:

$$\vec{r}_2 = (2)\hat{i} + (-1)\hat{j} + (0)\hat{k} + m\bigl(1\hat{i} + 1\hat{j} - 1\hat{k}\bigr).$$

Hence the coordinates of a general point on line 2 are

$$x_2 = 2 + m,\qquad y_2 = -1 + m,\qquad z_2 = 0 - m = -m.$$

If the two lines intersect, there must be values of $$l$$ and $$m$$ such that

$$x_1 = x_2,\quad y_1 = y_2,\quad z_1 = z_2.$$

Writing these three equations explicitly, we get

$$\begin{aligned} 1 + 2l &= 2 + m \quad &(1)\\[2pt] -1 &= -1 + m \quad &(2)\\[2pt] l &= -m \quad &(3) \end{aligned}$$

Now we solve this system step by step. From equation (2) we have

$$-1 = -1 + m \;\;\Longrightarrow\;\; m = 0.$$

Substituting $$m = 0$$ in equation (3) gives

$$l = -0 = 0.$$

Finally, we put $$l = 0$$ and $$m = 0$$ into equation (1):

$$1 + 2\cdot 0 = 2 + 0 \;\;\Longrightarrow\;\; 1 = 2,$$

which is clearly false. Hence the three equations cannot be satisfied simultaneously; there is no common point at any choice of $$l$$ and $$m$$.

Therefore the two given lines do not intersect for any values of $$l$$ and $$m$$.

Hence, the correct answer is Option A.