The statement $$(p \wedge q) \Rightarrow(p \wedge r)$$ is equivalent to :

Let the circumcentre of a triangle with vertices A(a, 3), B(b, 5) and C(a, b), ab > 0 be P(1,1). If the line AP intersects the line BC at the point Q$$\left(k_{1}, k_{2}\right)$$, then $$k_{1}+k_{2}$$ is equal to :

Let $$\hat{a}$$ and $$\hat{b}$$ be two unit vectors such that the angle between them is $$\frac{\pi}{4}$$. If $$\theta$$ is the angle between the vectors $$(\hat{a}+\hat{b})$$ and $$(\hat{a}+2 \hat{b}+2(\hat{a} \times \hat{b}))$$, then the value of $$164 \,\cos ^{2} \theta$$ is equal to :

If $$f(\alpha)=\int\limits_{1}^{\alpha} \frac{\log _{10} \mathrm{t}}{1+\mathrm{t}} \mathrm{dt}, \alpha>0$$, then $$f\left(\mathrm{e}^{3}\right)+f\left(\mathrm{e}^{-3}\right)$$ is equal to :