If $\bar{a}=\hat{i}+\hat{j}, \overline{\mathrm{~b}}=2 \hat{i}-\hat{\mathrm{k}}$ then the point of intersection of the lines $\overline{\mathrm{r}} \times \overline{\mathrm{a}}=\overline{\mathrm{b}} \times \overline{\mathrm{a}}$ and $\overline{\mathrm{r}} \times \overline{\mathrm{b}}=\overline{\mathrm{a}} \times \overline{\mathrm{b}}$ is

Finding the intersection point requires expressing both lines in standard vector form. The first equation $\overline{\mathrm{r}} \times \overline{\mathrm{a}} = \overline{\mathrm{b}} \times \overline{\mathrm{a}}$ simplifies to $(\overline{\mathrm{r}} - \overline{\mathrm{b}}) \times \overline{\mathrm{a}} = \overline{0}$, indicating $\overline{\mathrm{r}} - \overline{\mathrm{b}}$ is parallel to $\overline{\mathrm{a}}$. This yields the parametric form $\overline{\mathrm{r}} = \overline{\mathrm{b}} + \lambda \overline{\mathrm{a}}$.

The second equation $\overline{\mathrm{r}} \times \overline{\mathrm{b}} = \overline{\mathrm{a}} \times \overline{\mathrm{b}}$ similarly gives $(\overline{\mathrm{r}} - \overline{\mathrm{a}}) \times \overline{\mathrm{b}} = \overline{0}$, producing $\overline{\mathrm{r}} = \overline{\mathrm{a}} + \mu \overline{\mathrm{b}}$.

Equating both expressions for $\overline{\mathrm{r}}$ gives $\overline{\mathrm{b}} + \lambda \overline{\mathrm{a}} = \overline{\mathrm{a}} + \mu \overline{\mathrm{b}}$, which rearranges to $(\lambda - 1) \overline{\mathrm{a}} = (\mu - 1) \overline{\mathrm{b}}$.

With $\overline{\mathrm{a}} = \hat{i} + \hat{j}$ and $\overline{\mathrm{b}} = 2\hat{i} - \hat{\mathrm{k}}$, substitution yields:

$(\lambda - 1)\hat{i} + (\lambda - 1)\hat{j} = 2(\mu - 1)\hat{i} - (\mu - 1)\hat{\mathrm{k}}$

Comparing coefficients gives $\lambda - 1 = 0$ and $\mu - 1 = 0$, so $\lambda = \mu = 1$. The intersection point is $\overline{\mathrm{r}} = 3\hat{i} + \hat{j} - \hat{\mathrm{k}}$, corresponding to $(3, 1, -1)$.

The correct answer is $\boxed{\text{B}}$.