Question

$x^{2} + 4x + 3 - \alpha \leq 0$

and $x^{2} - 2x - (3 - 6\alpha) \leq 0$

$$\begin{matrix} \Rightarrow \ x & \ \in \lbrack 1 - \sqrt{4 - 6\alpha},1 + \sqrt{4 - 6\alpha}\rbrack \\ \alpha & \ = - 1\text{~satisfy the condition~} \\ \alpha & \ = \frac{2}{3}\text{~doesn't satisfy~} \end{matrix}$$

If

$$\begin{matrix} \alpha & \ \in \left( - 1,\frac{2}{3} \right) \\ \alpha + 1 & \ = 1 - \sqrt{4 - 6\alpha} \end{matrix}$$

$$\begin{matrix} & - 2 + \sqrt{\alpha + 1} & & = 1 - \sqrt{4 - 6\alpha} \\ & \Rightarrow & 7\alpha + 6 & = 6\sqrt{1 + \alpha} \Rightarrow \alpha \geq - \frac{6}{7} \\ \Rightarrow & & \Rightarrow 49\alpha^{2} + 48\alpha & = 0 \\ \Rightarrow & \alpha & & = 0,\frac{48}{49} \\ & & \alpha & = - \frac{48}{49}\text{~is rejected~} \\ & \text{~Hence,~} & \alpha & = 0, - 1 \end{matrix}$$