Answer
$\dfrac{-1}{\sqrt 3} i-\dfrac{1}{\sqrt 3}j-\dfrac{1}{\sqrt 3}k$
and
$(\dfrac{5}{2},\dfrac{7}{2},\dfrac{9}{2})$
Work Step by Step
Let us consider $P_1$ and $P_2$ as vectors and, a vector
$u=\overrightarrow{P_1P_2}=P_2-P_1=-i-j-k$
and $|u|=\sqrt{(1)^2+(-1)^2+(-1)^2}=\sqrt {3}$
The unit vector $\hat{\textbf{u}}$ can be calculated as: $\hat{\textbf{u}}=\dfrac{u}{|u|}$
Now, $\hat{\textbf{u}}=\dfrac{-i-j-k}{\sqrt 3}=\dfrac{-1}{\sqrt 3} i-\dfrac{1}{\sqrt 3}j-\dfrac{1}{\sqrt 3}k$
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, $(\dfrac{3+2}{2},\dfrac{4+3}{2},\dfrac{5+ 4}{2})=(\dfrac{5}{2},\dfrac{7}{2},\dfrac{9}{2})$
