Using Vector property finding the type of triangle

Question

If $\bar q , \bar p$, are two noncollinear and nonzero vectors such that $(b-c)\bar p × \bar q+(c-a)\bar p+(a-b)\bar q=0$ , where a, b, c are the length of the sides of a triangle, then the triangle is

(A) right angled (B) obtuse angled (C) equilateral (D) isosceles

I tried to square the term $(b-c)\bar p × \bar q=-(c-a)\bar p-(a-b)\bar q$ but it is getting more and more complicated

Answer 1

Hint: Show that $p,q$ and $p\times q$ form a basis of $\mathbb{R}^3$ to conclude that $a=b=c$.

Answer 2

Hint: $\vec{p},\vec{q}$ and $\vec{p}\times\vec{q}$ are linearly independent.

Answer 3

Multiply $(b-c)\bar p \times \bar q+(c-a)\bar p+(a-b)\bar q=0$ by $\bar p$ and $ \bar q$ to obtain a homogeneous linear system for $c-a$ and $a-b$ with only trivial solution.