$\vec{PP_1}+\vec{PP_2}+\vec{PP_3}=\frac {3}{2}\vec {PO} $. Let $ABC $ an equilateral  triangle and $P\in int (ABC) $. 
If $O $ is the center of gravity of  the triangle and $P_1, P_2, P_3$ are the projections of $P $ on the sides then 
$\vec{PP_1}+\vec{PP_2}+\vec{PP_3}=\frac {3}{2}\vec {PO} $.

Let 
$A_1C_2||AC $, $A_2B_1||AB $ and $B_2C_1||BC $ trough $P $
where $A_1,B_2\in [AB] $, $A_2, C_1\in [AC ]$ and $B_1, C_2\in [BC] $.
Then $PA_1AA_2$,  $PC_1CC_2$,  $PB_1BB_2$ are parallelograms and $A_1PB_2$, $A_2PC_1$, $B_1PC_2$ are equilateral triangles.
Now,
$\vec {PP_1}=\frac {\vec{PB_1}+\vec{PC_2}}{2}$
$\vec {PP_2}=\frac {\vec{PA_2}+\vec{PC_1}}{2}$
$\vec {PP_3}=\frac {\vec{PA_1}+\vec{PB_2}}{2}$.
But 
$\vec {PA_1}+\vec {PA_2}=\vec {PA} $
$\vec {PB_1}+\vec {PB_2}=\vec {PB} $
$\vec {PC_1}+\vec {PC_2}=\vec {PC} $.
So,
$\vec {PP_1}+\vec {PP_2}+\vec {PP_3}=\frac {\vec {PA}+\vec {PB}+\vec {PC}}{2}=\frac {3\vec {PO}}{2} $, q.e.d.
