1. **State the problem:** Evaluate the definite integral $$\int_0^1 3x \sqrt{1 - x} \, dx.$$\n\n2. **Rewrite the integral:** Note that $$\sqrt{1 - x} = (1 - x)^{1/2}.$$ So the integral becomes $$\int_0^1 3x (1 - x)^{1/2} \, dx.$$\n\n3. **Use substitution:** Let $$u = 1 - x,$$ then $$du = -dx,$$ and when $$x=0,$$ $$u=1,$$ and when $$x=1,$$ $$u=0.$$ Also, $$x = 1 - u.$$\n\n4. **Rewrite the integral in terms of $$u$$:**\n$$\int_0^1 3x (1 - x)^{1/2} \, dx = \int_1^0 3(1 - u) u^{1/2} (-du) = \int_0^1 3(1 - u) u^{1/2} \, du.$$\n\n5. **Expand the integrand:**\n$$3(1 - u) u^{1/2} = 3(u^{1/2} - u^{3/2}).$$\n\n6. **Split the integral:**\n$$\int_0^1 3(u^{1/2} - u^{3/2}) \, du = 3 \int_0^1 u^{1/2} \, du - 3 \int_0^1 u^{3/2} \, du.$$\n\n7. **Integrate each term:**\n- $$\int_0^1 u^{1/2} \, du = \left[ \frac{u^{3/2}}{3/2} \right]_0^1 = \frac{2}{3}.$$\n- $$\int_0^1 u^{3/2} \, du = \left[ \frac{u^{5/2}}{5/2} \right]_0^1 = \frac{2}{5}.$$\n\n8. **Calculate the value:**\n$$3 \times \frac{2}{3} - 3 \times \frac{2}{5} = 2 - \frac{6}{5} = \frac{10}{5} - \frac{6}{5} = \frac{4}{5}.$$\n\n**Final answer:** $$\boxed{\frac{4}{5}}.$$\n\nThis corresponds to option B.