Prove some properties of a plane determined by points P,Q,R

Let $P,Q,R$ be three points in $\mathbb{R}^3$ not all on the same line, and let $M$ be the plane they determine.

If $p,q,r \in \mathbb{R}$ with $p+q+r = 1$ prove that the point $pP + qQ + rR$ is on $M$ .
Prove that all of the points on $M$ are of the form $pP + qQ + rR$ for $p,q,r \in \mathbb{R}$ with $p+q+r = 1$ .

Proof. Since $M$ is the plane determined by the three points $P,Q,R$ we know $M$ is the set of points
$M = \{ P + s(Q-P) + t(R-P) \} \qquad s,t \in \mathbb{R}.$

Let $s = q, \ t = r$ . Then,

$\begin{align*} P + q(Q-P) + r(R-P) \in M && \implies && P - qP - rR + qQ + rR &\in M \\ && \implies && (1-q-r)P + qQ + rR &\in M \\ && \implies && pP + qQ + rR &\in M \end{align*}$

since $p+q+r = 1$ implies $p= 1 - q - r. \qquad \blacksquare$
Proof. Let $X$ be a point on $M$ . Then, we know there exist $s,t \in \mathbb{R}$ such that
$\begin{align*} && X &= P + s(Q-P) + t(R-P) \\ \implies && X &= (1-s-t)P + sQ + tR. \end{align*}$

But, $1-s-t + s+t = 1$ ; thus, $X = pP + qQ + rR$ for $p = 1-s-t, \ q = s, \ r = t. \qquad \blacksquare$

Stumbling Robot