Let $P=(a,b,c)$ and let $Q=(x,y,z)$. Then the line segment $PQ$ is the following set of points:
$$PQ=\{t\cdot(x,y,z)+(1-t)\cdot(a,b,c)\,\vert\,t\in[0,1]\}.$$
In other words, the points on the line segment $PQ$ are the points that can be expressed as
$$t\cdot(x,y,z)+(1-t)\cdot(a,b,c)$$
for some $t$ with $0\le t\le1$.
Edit: I was asked to give some justification of the above claim.
We have that $P=(a,b,c)$ and $Q=(x,y,z)$.
One way to parametrize the line $PQ$, is to note that points on the line are points of the form
$$P+t\cdot\vec{PQ}$$
where $t$ is a scalar. $\vec{PQ}$ is the vector going from $P$ to $Q$. Hence $\vec{PQ}$ can be represented as follows:
$$\begin{align*}
\vec{PQ} &= Q-P \\
&= (x,y,z)-(a,b,c) \\
&= (x-a,y-b,z-c)
\end{align*}$$
Hence points on the line $PQ$ can be represented as
$$\begin{align*}
P+t\cdot\vec{PQ} &= (a,b,c)+t\cdot(x-a,y-b,z-c) \\
&= (a,b,c)+t\cdot\left[((x,y,z)-(a,b,c)\right] \\
&= t\cdot(x,y,z)+(a,b,c)-t\cdot(a,b,c) \\
&= t\cdot(x,y,z)+(1-t)\cdot(a,b,c)
\end{align*}$$
for some scalar $t$. So we've established that every point on line $PQ$ can be written as
$$t\cdot(x,y,z)+(1-t)\cdot(a,b,c)$$
for some scalar $t$.
Now note the following: if $t=0$, then
$$t\cdot(x,y,z)+(1-t)\cdot(a,b,c)=(a,b,c)=P.$$
And if $t=1$, then
$$t\cdot(x,y,z)+(1-t)\cdot(a,b,c)=(x,y,z)=Q.$$
And if we let $0<t<1$, then $t\cdot(x,y,z)+(1-t)\cdot(a,b,c)$ will be a point between $P$ and $Q$.
It follows that the line segment $PQ$ consists of all points of the form
$$t\cdot(x,y,z)+(1-t)\cdot(a,b,c)$$
where $t$ is a scalar with $0\le t\le1$.
