Answer
$\dfrac{3 \sqrt 2}{2}\approx 2.12$
Work Step by Step
The parametric equations are:
$x=1+t; y=2-t; z=-1+2t$
or, $0=1(1+t) -1(2-t) +2(-1+2t)$
or, $1+t-2+t-2+4t=0$
or, $ 6t-3=0 \implies t=\dfrac{1}{2}$
Now, the distance formula is:
$D=\sqrt {(\dfrac{3}{2}-0)^2 +(\dfrac{3}{2}-0)^2 +(0-0)^2 }=\sqrt {\dfrac{9}{2}}$
So, $D=\dfrac{3}{\sqrt 2}\approx 2.12$