In each question two equations numbered (I) and (II) are given. You should solve both the equations and mark the appropriate answer.
I. $$3x^2 - 14x + 8 = 0$$
II. $$3y^2 - 20y + 12 = 0$$

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**Solution:**

1. Solve equation (I): $$3x^2 - 14x + 8 = 0$$

Using the quadratic formula:
$$x = \frac{14 \pm \sqrt{(-14)^2 - 4 	imes 3 	imes 8}}{2 	imes 3} = \frac{14 \pm \sqrt{196 - 96}}{6} = \frac{14 \pm \sqrt{100}}{6}$$

So,
$$x = \frac{14 \pm 10}{6}$$

- For the plus sign:
  $$x = \frac{14 + 10}{6} = \frac{24}{6} = 4$$

- For the minus sign:
  $$x = \frac{14 - 10}{6} = \frac{4}{6} = \frac{2}{3}$$

Thus, the roots of equation (I) are $x = 4$ and $x = \frac{2}{3}$.

2. Solve equation (II): $$3y^2 - 20y + 12 = 0$$

Using the quadratic formula:
$$y = \frac{20 \pm \sqrt{(-20)^2 - 4 	imes 3 	imes 12}}{2 	imes 3} = \frac{20 \pm \sqrt{400 - 144}}{6} = \frac{20 \pm \sqrt{256}}{6}$$

So,
$$y = \frac{20 \pm 16}{6}$$

- For the plus sign:
  $$y = \frac{20 + 16}{6} = \frac{36}{6} = 6$$

- For the minus sign:
  $$y = \frac{20 - 16}{6} = \frac{4}{6} = \frac{2}{3}$$

Thus, the roots of equation (II) are $y = 6$ and $y = \frac{2}{3}$.

3. Now, compare the roots of both equations:

- Common root: $\frac{2}{3}$
- Other roots: $x = 4$ and $y = 6$

Since $4 < 6$, for one pair $x < y$, and for the other pair $x = y$.

Therefore, no definite relation can be established between $x$ and $y$.

**Answer:** (a) If $x = y$ or no relation can be established

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