Let $u$, $v$, and $w$ be any three non-coplanar vectors emanating from the origin of the three-dimensional Euclidean space. Let $x = au + bv + cw$, where $a$, $b$, $c$ are scalars. Then what is the necessary and sufficient condition (on $a$, $b$, and $c$) for the head of $x$ to lie
(i) in the interior
(ii) on the boundary
of the triangle determined by $u$, $v$, and $w$?
My effort:
The equation of the plane determined by the head points of $u$, $v$, and $w$ is $$ (y - u) \cdot (y-v) \times (y-w) = 0.$$ Am I right?
So in either case we must have $$ (x - u) \cdot (x-v) \times (x-w) = 0.$$ Or, $$ [ (a-1) u + bv + cw ] \cdot [ au + (b-1)v + cw ] \times [ au + bv + (c-1)w ] = 0.$$ The last equation simplifies to $$ (a + b + c -1 ) u \cdot v \times w = 0.$$ But as $u$, $v$, and $w$ are non-coplanar, so $$ u \cdot v \times w \neq 0,$$ and therefore $$a+b+c=1.$$ Am I right?
What next?
I am not sure the rigor you need, but I can try:
Consider the plane spanned by $u$ and $v$ passing by the origin of coordinates. A plane splits the space in two parts. Now, $cw$ with $c>0$ is in one of the parts, obviously, the same the triangle is. With $c<0$ $cw$ points to the other part, so, $au+bv+cw$ with $c<0$ is in the triangle's exterior. Name the half space the triangle is as "positive". Analogously for the other two planes ($u$ with $w$ and $v$ with $w$). So $a,b,c\ge 0$ is a necessary condition. It is a sufficient condition too: the triangle is into the intersection of the three positive half spaces considered above
One of the sides of the triangle is into the plane $mu+nv=0$, so if $au+bv=0$ is in that side. $c=0$ is a necessary and sufficient condition for the point to be in that side. Analogously for the other sides.