Answer
$\dfrac{3}{7} i-\dfrac{6}{7}j+\dfrac{2}{7}k$ and $(\dfrac{5}{2},1,6)$
Work Step by Step
Let us consider $P_1$ and $P_2$ as vectors and, a vector
$u=\overrightarrow{P_1P_2}=P_2-P_1=(4-1)i+(-2-4)j+(7-5)k=3i-6j+2k$
and $|u|=\sqrt{(3)^2+(-6)^2+(2)^2}=\sqrt {49}=7$
The unit vector $\hat{\textbf{u}}$ can be calculated as: $\hat{\textbf{u}}=\dfrac{u}{|u|}$
Now, $\hat{\textbf{u}}=\dfrac{3i-6j+2k}{7}=\dfrac{3}{7} i-\dfrac{6}{7}j+\dfrac{2}{7}k$
The mid-point can be found as: $(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2},\dfrac{z_1+z_2}{2})$
Thus, $(\dfrac{1+4}{2},\dfrac{4+(-2)}{2},\dfrac{5+7}{2})=(\dfrac{5}{2},1,6)$
