Given points lie on the line $y=x-1$ and the parabola $y=x^2+x+1$
Thus, the minimum distance occurs along the common normal
$\therefore$ At $B, \frac{dy}{dx}=1$

or $2x+1=1$
$\Rightarrow x=0$
$\therefore B$ is $\left(0,1\right)$
Hence, the shortest distance is $D=\frac{\left|0-1-1\right|}{\sqrt{2}}=\sqrt{2}$ units