The given equations are, 2x−y+3z−4=0⇒2x−y+3z=4
Multiplying equation by 3 we get,
⇒6x−3y+9z=12⋯(i)
6x−3y+9z+13=0⇒6x−3y+9z=−13⋯(ii)
Comparing the given equations with ax+by+cz=d1 and ax+by+cz=d2 we get,
a=6,b=−3,c=9,d1=12,d2=−13
Distance between plane 1 and 2 =|d1−d2|√a2+b2+c2
D=|12+13|√36+9+81=25√126
∴126D2=625