The correct option is C 51x+15y−50z+173=0
Equation of plane through intersection of the two planes is,
P:(x+2y+3z−4)+λ(2x+y−z+5)=0 …(1)
→nP=(1+2λ)^i+(2+λ)^j+(3−λ)^k
Direction ratios of its normal are 2λ+1,λ+2,3−λ
Again, consider the plane 5x+3y+6z+8=0 …(2)
Direction ratios of its normal are 5,3,6
Since plane (1) is perpendicular to plane (2),
nP⋅→n=0⇒5(1+2λ)+3(2+λ)+6(3−λ)=0
⇒7λ+29=0⇒λ=−297
Putting λ=−297 in equation (1), we get
(x+2y+3z−4)−297(2x+y−z+5)=0⇒7(x+2y+3z−4)−29(2x+y−z+5)=0⇒7x+14y+21z−28−58x−29y+29z−145=0⇒51x+15y−50z+173=0