The sum of squares of all the real solutions of the equation

$\log _{(x+1)}\left(2 x^2+5 x+3\right)=4-\log _{(2 x+3)}\left(x^2+2 x+1\right)$ is equal to $\_\_\_\_$ .

If $\int\limits_{\pi / 6}^{\pi / 4}\left(\cot \left(x-\frac{\pi}{3}\right) \cot \left(x+\frac{\pi}{3}\right)+1\right) d x=\alpha \log _{\mathrm{e}}(\sqrt{3}-1)$, then $9 \alpha^2$ is equal to $\_\_\_\_$ .

Let a line $L_1$ pass through the origin and be perpendicular to the lines

$\mathrm{L}_2: \overrightarrow{\mathrm{r}}=(3+\mathrm{t}) \hat{i}+(2 \mathrm{t}-1) \hat{j}+(2 \mathrm{t}+4) \hat{k}$ and

$\mathrm{L}_3: \overrightarrow{\mathrm{r}}=(3+2 \mathrm{~s}) \hat{i}+(3+2 \mathrm{~s}) \hat{j}+(2+\mathrm{s}) \hat{k}, \mathrm{t}, \mathrm{s} \in \mathbf{R}$.

If $(a, b, c), a \in \mathbf{Z}$, is the point on $\mathrm{L}_3$ at a distance of $\sqrt{17}$ from the point of intersection of $\mathrm{L}_1$ and $\mathrm{L}_2$, then $(\mathrm{a}+\mathrm{b}+\mathrm{c})^2$ is equal to $\_\_\_\_$ .

Consider the circle C : $x^2+y^2-6 x-8 y-11=0$. Let a variable chord AB of the circle C subtend a right angle at the origin. If the locus of the foot of the perpendicular drawn from the origin on the chord AB is the circle $x^2+y^2-\alpha x-\beta y-\gamma=0$, then $\alpha+\beta+2 \gamma$ is equal to $\_\_\_\_$ .