Given that the roots of $3x^2 - x + 2 = 0$ are $\alpha$ and $\beta$, find a quadratic equation whose roots are $\frac{\alpha + 1}{\beta}$ and $\frac{\beta + 1}{\alpha}$.

1. Find the sum and product of the original roots $\alpha$ and $\beta$.

From $3x^2 - x + 2 = 0$:
$\alpha + \beta = -\frac{-1}{3} = \frac{1}{3}$
$\alpha\beta = \frac{2}{3}$

2. Express the sum of the new roots in terms of $\alpha + \beta$ and $\alpha\beta$.

Sum of new roots $= \frac{\alpha + 1}{\beta} + \frac{\beta + 1}{\alpha}$
$= \frac{\alpha(\alpha + 1) + \beta(\beta + 1)}{\alpha\beta}$
$= \frac{\alpha^2 + \alpha + \beta^2 + \beta}{\alpha\beta}$
$= \frac{(\alpha + \beta)^2 - 2\alpha\beta + (\alpha + \beta)}{\alpha\beta}$

3. Calculate the sum of the new roots.

Sum $= \frac{(\frac{1}{3})^2 - 2(\frac{2}{3}) + \frac{1}{3}}{\frac{2}{3}}$
$= \frac{\frac{1}{9} - \frac{4}{3} + \frac{1}{3}}{\frac{2}{3}}$
$= \frac{-\frac{8}{9}}{\frac{2}{3}}$
$= -\frac{4}{3}$

4. Express and calculate the product of the new roots.

Product of new roots $= \left(\frac{\alpha + 1}{\beta}\right)\left(\frac{\beta + 1}{\alpha}\right)$
$= \frac{\alpha\beta + \alpha + \beta + 1}{\alpha\beta}$
$= \frac{\frac{2}{3} + \frac{1}{3} + 1}{\frac{2}{3}}$
$= \frac{2}{\frac{2}{3}}$
$= 3$

5. Form the new quadratic equation.

$x^2 - (\text{Sum of roots})x + (\text{Product of roots}) = 0$
$x^2 - \left(-\frac{4}{3}\right)x + 3 = 0$
$x^2 + \frac{4}{3}x + 3 = 0$
$3x^2 + 4x + 9 = 0$