1. **State the problem:** Solve the inequality $$\frac{(2-x)^3 (x-21)^6}{(2x-3)^2} \ge 0.$$\n\n2. **Identify critical points:** The numerator is zero at $x=2$ and $x=21$. The denominator is zero at $x=\frac{3}{2}$. These points divide the real line into intervals to test.\n\n3. **Analyze the sign of each factor:**\n- $(2-x)^3$ changes sign at $x=2$ and has odd multiplicity (3), so it changes sign there.\n- $(x-21)^6$ is always nonnegative and zero at $x=21$ (even multiplicity 6), so it does not change sign.\n- $(2x-3)^2$ is always positive except zero at $x=\frac{3}{2}$ (even multiplicity 2), so denominator is zero and undefined at $x=\frac{3}{2}$.\n\n4. **Determine sign on intervals:**\n- For $x < \frac{3}{2}$: $(2-x)^3 > 0$ (since $2-x > 0$), $(x-21)^6 > 0$, denominator positive, so expression $>0$.\n- At $x=\frac{3}{2}$: denominator zero, expression undefined, exclude this point.\n- For $\frac{3}{2} < x < 2$: $(2-x)^3 < 0$ (since $2-x < 0$), numerator negative, denominator positive, expression $<0$.\n- At $x=2$: numerator zero, expression equals zero, included since inequality is $\ge 0$.\n- For $2 < x < 21$: $(2-x)^3 < 0$, $(x-21)^6 > 0$, numerator negative, denominator positive, expression $<0$.\n- At $x=21$: numerator zero, expression equals zero, included.\n- For $x > 21$: $(2-x)^3 < 0$, $(x-21)^6 > 0$, numerator negative, denominator positive, expression $<0$.\n\n5. **Summary of solution:**\n- Expression positive on $(-\infty, \frac{3}{2})$.\n- Undefined at $x=\frac{3}{2}$.\n- Negative on $(\frac{3}{2}, 2)$.\n- Zero at $x=2$.\n- Negative on $(2, 21)$.\n- Zero at $x=21$.\n- Negative for $x > 21$.\n\n6. **Final solution set:** $$(-\infty, \frac{3}{2}) \cup \{2\} \cup \{21\}.$$\n\n7. **Check answer choices:** None exactly match this set, but option (A) is $(-\infty, \frac{3}{2}) \cup (\frac{3}{2}, 2] \cup \{21\}$ which incorrectly includes $(\frac{3}{2}, 2]$ where expression is negative. Option (C) is the same as (A). Option (B) and (D) include intervals where expression is negative or undefined.\n\nTherefore, the correct solution is $$(-\infty, \frac{3}{2}) \cup \{2\} \cup \{21\}$$ which is not exactly listed but closest to (A) and (C) except they wrongly include $(\frac{3}{2}, 2]$.\n\n**Note:** Since the problem asks to solve and select from given options, the best match is (A) or (C) ignoring the incorrect inclusion of $(\frac{3}{2}, 2]$.\n