The following is the probability distribution of X

$$ \begin{array}{|c|c|c|c|c|} \hline \mathrm{X} & 0 & 1 & 2 & 3 \\ \hline \mathrm{P}(\mathrm{X}=x) & \frac{1+\mathrm{p}}{5} & \frac{2-2 \mathrm{p}}{5} & \frac{2-\mathrm{p}}{5} & \frac{2 \mathrm{p}}{5} \\ \hline \end{array} $$

$$ \text { For a minimum value of } p \text {, the value of } 5 E(X) \text { is } $$

Let $\overline{\mathrm{a}}$ and $\overline{\mathrm{c}}$ be unit vectors at an angle $\frac{\pi}{3}$ with each other. If $(\overline{\mathrm{a}} \times(\overline{\mathrm{b}} \times \overline{\mathrm{c}})) \cdot(\overline{\mathrm{a}} \times \overline{\mathrm{c}})=5$, then $[\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{c}}]=$

A random variable X takes values $0,1,2,3$, ........ with probabilities. $\mathrm{P}(\mathrm{X}=x)=\mathrm{k}(x+1)\left(\frac{1}{2}\right)^x, \mathrm{k}$ is a constant, then $P(X=1)=$

The area inside the parabola $y^2=4 \mathrm{a} x$, between the lines $x=\mathrm{a}$ and $x=4 \mathrm{a}$ is equal to