1. **State the problem:** We need to find the value of $x$ such that quadrilateral ABCD is a parallelogram. 2. **Given:** Angles at vertices A and C along diagonal AC are $(12x + 1)^\circ$ and $(14x - 5)^\circ$ respectively. 3. **Key property:** In a parallelogram, opposite angles are equal, and consecutive angles are supplementary. Here, since ABCD is a parallelogram, angles A and C are opposite angles and must be equal. 4. **Set up the equation:** $$12x + 1 = 14x - 5$$ 5. **Solve for $x$:** $$12x + 1 = 14x - 5$$ $$1 + 5 = 14x - 12x$$ $$6 = 2x$$ $$x = \frac{6}{2}$$ $$x = 3$$ 6. **Conclusion:** The value of $x$ must be $3$ for ABCD to be a parallelogram.