Consider the lines x = y = z and the line 2x + y + z –1 = 0 = 3x + y + 2z – 2 is

Any plane through the second line is 2x + y + z –1 + k (3x + y + 2z –2) = 0If this is parallel to the first line, then  (2+3k)+(1+k)+(1+2k)=0⇒k=−23⇒Plane is 2x+y+z−1−23(3x+y+2z−2)=0 or y−z+1=0. The required SD must be distance of this plane from any point on the line x=y=z say (1, 1, 1) ⇒ SD=|1−1+1|02+12+(−1)2=12