Vectorial height of a triangle 
I personally couldn't figure out how $\overrightarrow{\mathrm{h}}_1$ was derived. Any help would be very much appreciated.
$A$ part is understood which is the area of the triangle.
Extended question
Further how to find a point on $\overrightarrow{\ell}_2$ when given ${h_1}$. I have commented a method but not sure whether it's correct or not.
 A: The prime “ ‘ “ notation for magnitudes is just a complication. I have not used it. Also the “^” notation for vector is misleading because it usually is associated with a vector of magnitude 1. I prefer the regular vector notation.
You have a mistake for area. The correct formula is:
Area $A=\frac{\vec{n}\cdot (\vec {l_2}\times \vec {l_3})}{2}$
This formula of area accounts for:

*

*$\vec n$ is the normal vector to the plane of the triangle and its magnitude $||\vec n||=n=1$

*Furthermore, since $\vec{l_2}\times \vec{l_3}$ results in a vector parallel to $\vec n$, then the dot product is needed to have the resulting area A as a number instead of a vector.

*You can have the cross product of any pair of different  vectors $\vec{l_1}$, $\vec{l_2}$, $\vec{l_3}$ in the formula of the area: $A=\frac{\vec{n}\cdot (\vec {l_2}\times \vec {l_3})}{2}=\frac{\vec{n}\cdot (\vec {l_3}\times \vec {l_1})}{2}=\frac{\vec{n}\cdot (\vec {l_1}\times \vec {l_2})}{2}$
Then the area can be expressed as $\frac{base \cdot height}{2}$ which in terms of vectors, given that base and height are perpendicular:
$A=\frac{\vec n\cdot(\vec{h_1}\times \vec{l_1})}{2}=\frac{n \cdot h_1 \cdot l_1 \cdot sin(90^ \circ)}{2}=\frac{1\cdot h_1 \cdot l_1 \cdot 1}{2}\Rightarrow h_1=\frac{2A}{l_1}=\frac{2A}{l_1}\cdot\vec{l_1}\times \vec n$
At the last equality I’ve considered that $\vec{l_1}\times \vec n=||\vec{l_1}||\cdot||\vec n || \cdot sin(\vec{l_1},\vec n)=l_1 \cdot 1\cdot sin(90^ \circ)=l_1$
