Question 1209738
Here's how to solve this problem:

1. **Vieta's Formulas:**
   For a cubic equation of the form ax³ + bx² + cx + d = 0, Vieta's formulas relate the coefficients to sums and products of the roots. In our case, 3x³ - 4x² - 2x + 12 = 0, so a = 3, b = -4, c = -2, and d = 12. The formulas we'll need are:

   * rst = -d/a = -12/3 = -4
   * rs + rt + st = c/a = -2/3

2. **Rewrite the Expression:**
   We want to compute:

   (rs/t²) + (rt/s²) + (st/r²) = (r³s³ + r³t³ + s³t³) / (r²s²t²)

3. **Simplify using Vieta's Formulas:**
   We know that r²s²t² = (rst)² = (-4)² = 16.  So, we need to find r³s³ + r³t³ + s³t³.

4. **Key Identity:**
   Recall the identity:  A³ + B³ + C³ - 3ABC = (A + B + C)(A² + B² + C² - AB - BC - CA)
   Let A = rs, B = rt, and C = st.  Then:

   (rs)³ + (rt)³ + (st)³ - 3(rst)² = (rs + rt + st)[(rs)² + (rt)² + (st)² - rs*rt - rs*st - rt*st]

   We have rs + rt + st = -2/3 and rst = -4.  Substituting these values:

   r³s³ + r³t³ + s³t³ - 3(-4)² = (-2/3)[(rs)² + (rt)² + (st)² - r²st - rs²t - rst²]
   r³s³ + r³t³ + s³t³ - 48 = (-2/3)[(rs)² + (rt)² + (st)² - rt(rs + st + rt)]
   r³s³ + r³t³ + s³t³ - 48 = (-2/3)[(rs)² + (rt)² + (st)² - (-4)(-2/3)]
   r³s³ + r³t³ + s³t³ - 48 = (-2/3)[(rs)² + (rt)² + (st)² - 8/3]

   We also know that (rs + rt + st)² = (rs)² + (rt)² + (st)² + 2rst(r + s + t).
   (-2/3)² = (rs)² + (rt)² + (st)² + 2(-4)(4/3)
   4/9 = (rs)² + (rt)² + (st)² - 32/3
   (rs)² + (rt)² + (st)² = 4/9 + 96/9 = 100/9

   Now plug this back into the equation:
   r³s³ + r³t³ + s³t³ - 48 = (-2/3)[100/9 - 8/3]
   r³s³ + r³t³ + s³t³ - 48 = (-2/3)[100/9 - 24/9]
   r³s³ + r³t³ + s³t³ - 48 = (-2/3)(76/9)
   r³s³ + r³t³ + s³t³ = 48 - 152/27
   r³s³ + r³t³ + s³t³ = (1296 - 152)/27 = 1144/27

5. **Final Calculation:**

   (r³s³ + r³t³ + s³t³) / (r²s²t²) = (1144/27) / 16 = 1144 / (27 * 16) = 1144 / 432 = 143/54

Final Answer: The final answer is $\boxed{143/54}$