If the external bisectors of the angles of a triangle intersect the opposite sides at points P,Q,R, show P,Q,R are collinear. I'm required to solve this problem using vector algebra, but I cannot visualise the problem. 
If the external bisectors of the angles of a triangle intersect the opposite sides at points P,Q,R, show that P,Q,R are collinear.

I feel you man! Solving this one with linear algebra is so awkward. This is not a problem one should be solving by calculating crazy vectors. I'll show you another way. 
Let $ABC$ be the triangle in question and let $l_C$ be the external bisector of angle $\angle BCA$. Analogously let $l_A$ and $l_B$ be the external bisectors of angles $\angle CAB$ and $\angle ABC$ respectively.    
Next, let $L_C$ be the intersection point of the bisector $l_C$ with the line $AB,$ i.e. $L_C = l_C \cap AB$. Analogously, let $L_B = l_B \cap CA$ and $L_A = l_A \cap BC$. These three points $L_A, L_B, L_C$ are the ones you need to show are collinear.
In order to prove the latter fact, draw the three intersection points between the three pairs of external angle bisectors $A_1 = l_B \cap l_C, \,\, B_1 = l_C \cap l_A$ and $C_1 = l_A \cap l_B$. Now, since $A_1$ is the intersection point of the two external bisectors through vertices $B$ and $C$, the internal bisector through vertex $A$ should also pass through $A_1$. Therefore, the line $AA_1$ is the internal bisector of angle $\angle CAB$. Analogouly, $BB_1$ is the internal bisector of $\angle ABC$ and $CC_1$ is the internal bisector of $\angle BCA$. But as we know, the internal bisectors $AA_1, BB_1$ and $CC_1$ of triangle $ABC$ intersect at a common point (the incenter of $ABC$), so the two triangles $ABC$ and $A_1B_1C_1$ are perspective, i.e. they satisfy Desargue's Theorem. Therefore, since $l_A = B_1C_1, \,\, l_B = C_1A_1$ and $l_C = A_1B_1$ by construction, the three points $L_C = l_C \cap AB = A_1B_1 \cap AB, \,\, L_B = l_B \cap CA = C_1A_1 \cap CA$ and $L_A = l_A \cap BC = B_1C_1 \cap BC$ lie on a common line.    
