Let $\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}$ be a twice differentiable function such that $f(2)=1$. If $\mathrm{F}(\mathrm{x})=\mathrm{x} f(\mathrm{x})$ for all $\mathrm{x} \in \mathrm{R}$, $\int\limits_0^2 x F^{\prime}(x) d x=6$ and $\int\limits_0^2 x^2 F^{\prime \prime}(x) d x=40$, then $F^{\prime}(2)+\int\limits_0^2 F(x) d x$ is equal to :

The square of the distance of the point $ \left( \frac{15}{7}, \frac{32}{7}, 7 \right) $ from the line $ \frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7} $ in the direction of the vector $ \hat{i} + 4\hat{j} + 7\hat{k} $ is:

Two equal sides of an isosceles triangle are along $ -x + 2y = 4 $ and $ x + y = 4 $. If $ m $ is the slope of its third side, then the sum, of all possible distinct values of $ m $, is: