$$\text { The integral } \int \frac{\left(x^8-x^2\right) \mathrm{d} x}{\left(x^{12}+3 x^6+1\right) \tan ^{-1}\left(x^3+\frac{1}{x^3}\right)} \text { is equal to : }$$

The values of $$\alpha$$, for which $$\left|\begin{array}{ccc}1 & \frac{3}{2} & \alpha+\frac{3}{2} \\ 1 & \frac{1}{3} & \alpha+\frac{1}{3} \\ 2 \alpha+3 & 3 \alpha+1 & 0\end{array}\right|=0$$, lie in the interval

For $$0 < \mathrm{a} < 1$$, the value of the integral $$\int_\limits0^\pi \frac{\mathrm{d} x}{1-2 \mathrm{a} \cos x+\mathrm{a}^2}$$ is :

The position vectors of the vertices $$\mathrm{A}, \mathrm{B}$$ and $$\mathrm{C}$$ of a triangle are $$2 \hat{i}-3 \hat{j}+3 \hat{k}, 2 \hat{i}+2 \hat{j}+3 \hat{k}$$ and $$-\hat{i}+\hat{j}+3 \hat{k}$$ respectively. Let $$l$$ denotes the length of the angle bisector $$\mathrm{AD}$$ of $$\angle \mathrm{BAC}$$ where $$\mathrm{D}$$ is on the line segment $$\mathrm{BC}$$, then $$2 l^2$$ equals :