Assume, $Q$ be the midpoint of $\left(1,2,3\right)$ & $\left(-\frac{7}{3},-\frac{4}{3},-\frac{1}{3}\right)$

So, coordinate of 

$Q=\left(\frac{1-{\displaystyle \frac{7}{3}}}{2},\frac{2-{\displaystyle \frac{4}{3}}}{2},\frac{3-{\displaystyle \frac{1}{3}}}{2}\right)=\left(\frac{-2}{3},\frac{1}{3},\frac{4}{3}\right)$

The direction ratio of normal to the plane is 

$1+\frac{7}{3}, 2+\frac{4}{3}, 3+\frac{1}{3}=\frac{10}{3},\frac{10}{3},\frac{10}{3} \Rightarrow 1,1,1$

So, equation of plane is $x+y+z=1$.

So, $\left(1,-1,1\right)$ lies on this plane.