Prove that mvec a-nvec b , pvec b-mvec c and nvec c-pvec a…

Prove that $m\vec a-n\vec b$, $p\vec b-m\vec c$ and $n\vec c-p\vec a$ are always coplanar.

Explanation of Method

The vectors ma−nb, pb−mc and nc−pa are always coplanar.

Step 1: A vanishing linear combination

Let $\vec u=m\vec a-n\vec b$, $\vec v=p\vec b-m\vec c$, $\vec w=n\vec c-p\vec a$. Multiply them by $p$, $n$, $m$ respectively: $p\vec u+n\vec v+m\vec w=pm\,\vec a-pn\,\vec b+np\,\vec b-nm\,\vec c+mn\,\vec c-mp\,\vec a=\vec 0$.

Step 2: Conclusion

If $(p,n,m)\neq(0,0,0)$, this is a non-trivial linear dependence, so $\vec u,\vec v,\vec w$ are coplanar.
If $m=n=p=0$, all three vectors are $\vec 0$, which is trivially coplanar.

Equivalently, $[\vec u,\vec v,\vec w]=0$. For example, if $p\neq0$ then $\vec u=-\tfrac1p(n\vec v+m\vec w)$, and the triple product vanishes.

\[p\,(m\vec a-n\vec b)+n\,(p\vec b-m\vec c)+m\,(n\vec c-p\vec a)=\vec 0\ \Longrightarrow\ \text{always coplanar}\qquad\blacksquare\]

Result

\[\boxed{\,p\,(m\vec a-n\vec b)+n\,(p\vec b-m\vec c)+m\,(n\vec c-p\vec a)=\vec 0\ \Longrightarrow\ \text{always coplanar}\qquad\blacksquare\,}\]

Remark

Correction. The source claims $\vec u+\vec v+\vec w=\vec 0$ because the coefficients $(m-p)+(p-n)+(n-m)$ add up to $0$. That does not follow: \[\vec u+\vec v+\vec w=(m-p)\vec a+(p-n)\vec b+(n-m)\vec c,\] which is not $\vec 0$ in general (e.g. $m=1$, $n=p=0$ gives $\vec a-\vec c$). The correct weights are $p,n,m$, not $1,1,1$.