The shortest distance between the lines $2x + y + z - 1= 0, 3x + y + 2z - 2=0 $ and $ x = y = z, $ is

$\frac{1}{\sqrt{2}}$

The shortest distance between the given lines is equal to the length of the perpendicular from any point on $\frac{x-0}{1}=\frac{y-0}{1}=\frac{z-0}{1}$ to the plane through the line $2x + y + z - 1 = 10, 3x + y + 2z - 2 = 0 $ and parallel to $\frac{x-0}{1}=\frac{y-0}{1}=\frac{z-0}{1}$.

The equation of any plane through the line 2x + y + z - 1= 10, 3x + y + 2z - 2 = 0 is

$(2x + y + z - 1) +λ (3x + y + 2z - 2) = 0 $

or, $x (3 λ + 2) + y ( λ + 1) + z (2λ + 1) - 2λ = 0 $ .....(i)

If it is parallel to $\frac{x-0}{1}=\frac{y-0}{1}=\frac{z-0}{1}$, then

$3λ + 2 + λ + 1 + 2 λ + 1 = 0 ⇒ λ = \frac{-2}{3}$

Putting $ λ = \frac{-2}{3}$ in (i), we obtain -y + z - 1= 0.

The distance of this plane from any point on $\frac{x-1}{1}=\frac{y-0}{1}=\frac{z-0}{1}$ is

$ d = \left|\frac{-0+0-1}{\sqrt{1+1}}\right|= \frac{1}{\sqrt{2}}$