A line having direction ratios $1,-4,2$ intersects the lines $\frac{x-7}{3}=\frac{y-1}{-1}=\frac{z+2}{1}$ and $\frac{x}{2}=\frac{y-7}{3}=\frac{z}{1}$ at the points $A$ and $B$ resp., then co-ordinates of points A and B are

$\int \frac{x^2-4}{x^4+9 x^2+16} \mathrm{dx}=\tan ^{-1}(\mathrm{f}(x))+\mathrm{c}$ (where c is a constant of integration), then value of $f(2)$ is

If $\overline{\mathrm{a}}=\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$ and $\overline{\mathrm{b}}=2 \hat{i}+3 \hat{\mathrm{j}}-\hat{\mathrm{k}}$ are two vectors, then the angle between the vectors $3 \overline{\mathrm{a}}+5 \overline{\mathrm{~b}}$ and $5 \overline{\mathrm{a}}+3 \overline{\mathrm{~b}}$ is

Let $A=\left[\begin{array}{cc}1 & 2 \\ -1 & 4\end{array}\right]$ and $A^{-1}=\alpha \mathrm{I}+\beta \mathrm{A}, \alpha, \beta \in \mathbb{R}$, I is the identity matrix of order 2 , then $4(\alpha-\beta)$ is