The distance of the plane $\overline{\mathrm{r}}=(\hat{i}-\hat{\mathrm{j}})+\lambda(\hat{i}+\hat{\mathrm{j}}+\hat{\mathrm{k}})+\mu(\hat{i}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})$ from the origin is

The plane's distance from the origin is determined by converting its vector equation to standard form and applying the distance formula.

The plane equation $\overline{\mathrm{r}} = (\hat{i}-\hat{\mathrm{j}}) + \lambda(\hat{i}+\hat{\mathrm{j}}+\hat{\mathrm{k}}) + \mu(\hat{i}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})$ indicates a normal vector $\overline{\mathrm{n}} = \overline{\mathrm{b}} \times \overline{\mathrm{c}} = 5\hat{i} - 2\hat{\mathrm{j}} - 3\hat{\mathrm{k}}$. Since point $\hat{i}-\hat{\mathrm{j}}$ lies on the plane, the scalar constant evaluates to $d = \overline{\mathrm{a}} \cdot \overline{\mathrm{n}} = 7$.

The Cartesian form becomes $5x - 2y - 3z = 7$. The distance from the origin to this plane is given by $\frac{|7|}{\sqrt{5^2 + (-2)^2 + (-3)^2}} = \frac{7}{\sqrt{38}}$ units.

Final answer: $\boxed{\frac{7}{\sqrt{38}}}$ units