1. **State the problem:** We need to solve for $D$ where $$D = 2 \times \sqrt{\frac{3}{2}} - 3 \times \sqrt{\frac{2}{3}} - 1.$$\n\n2. **Recall the properties of square roots and simplification:**\n- $\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$.\n- We can simplify square roots by factoring out perfect squares.\n\n3. **Rewrite each square root term:**\n$$2 \times \sqrt{\frac{3}{2}} = 2 \times \frac{\sqrt{3}}{\sqrt{2}} = \frac{2\sqrt{3}}{\sqrt{2}}$$\n$$3 \times \sqrt{\frac{2}{3}} = 3 \times \frac{\sqrt{2}}{\sqrt{3}} = \frac{3\sqrt{2}}{\sqrt{3}}$$\n\n4. **Rationalize the denominators:**\nMultiply numerator and denominator by the denominator's conjugate to remove the square root from the denominator.\n\nFor the first term:\n$$\frac{2\sqrt{3}}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{2\sqrt{3} \times \sqrt{2}}{2} = \frac{2\sqrt{6}}{2}$$\nSimplify numerator and denominator:\n$$= \cancel{2} \sqrt{6} / \cancel{2} = \sqrt{6}$$\n\nFor the second term:\n$$\frac{3\sqrt{2}}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{3\sqrt{2} \times \sqrt{3}}{3} = \frac{3\sqrt{6}}{3}$$\nSimplify numerator and denominator:\n$$= \cancel{3} \sqrt{6} / \cancel{3} = \sqrt{6}$$\n\n5. **Substitute back into the expression for $D$:**\n$$D = \sqrt{6} - \sqrt{6} - 1$$\n\n6. **Simplify:**\n$$\sqrt{6} - \sqrt{6} = 0$$\nSo, $$D = 0 - 1 = -1$$\n\n**Final answer:** $$D = -1$$