Question

Sol. [Key Concept: Two points lie on opposite sides of a line when the line expression takes opposite signs there; solve (value at $P$)(value at $Q$)$< 0$ for $\alpha$.]

Put each point in $f = 2x + 3y - 6$:

$f(P_{1}) = 2\alpha + 3(2 + \alpha) - 6 = 5\alpha$,

$f(P_{2}) = 3\alpha + 3\alpha^{2} - 6 = 3(\alpha + 2)(\alpha - 1)$.

Opposite sides

$$\Rightarrow \ f\left( P_{1} \right)f\left( P_{2} \right) < 0:\ \ $$

$$15\alpha(\alpha + 2)(\alpha - 1) < 0$$

$\Rightarrow \ \alpha \in ( - \infty, - 2) \cup (0,1)$ → (a), (c).