The equation of the plane containing the straight line x2=y3=z4 and perpendicular to the plane containing the  straight lines x3=y4=z2 and x4=y2=z3 is px+qy+r=0, then p+q+r is

Let the direction ratios of the plane containing lines x3=y4=z2 and is x4=y2=z3 be a,b,c∴  3a+4b+2c=0 and   4 a+2 b+3 c=0∴  a12-4=b8-9=c6-16      ⇒a8=b-1=c-10∴ Direction ratio of plane =-8,1,10  Let the direction ratio of required plane is ⟨l,m,n⟩ Then -8l+m+10n=0 and 2l+3m+4n=0     ∴l-26=m52=n-26∴   D.R.s are <1,-2,1>∴   Equation of plane: x-2y+z=0⇒px+qy+r=x-2y+z⇒p=1,q=-2,r=1 Hence, p+q+r=1-2+1=0