The shortest distance between the lines 

$2x+y+z-1=0=3x+y+2z-2,\text{ and }x=y=z,\text{ is }$

: Any plane passing through first line is  $2x+y+z-1+\lambda (3x+y+2z-2)=0$

Line x = y = z is passing through the point 0(0, 0; 0).Required shortest distance = distance of O from the

member plane of above family which is parallel to the

line x = y = z

If plane is parallel to the line

$(2+3\lambda )1+(1+\lambda )1+(1+2\lambda )1=0\Rightarrow \lambda =-\frac{2}{3}$

Equation of plane is

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

Its distance from $(0,0,0)\text{ is }\frac{1}{\sqrt{2}}$