The correct option is B (2,5)
log1/3(x−1)+log1/3(x+1)+log√3(5−x) is valid when x−1>0,x+1>0,5−x>0
⇒1<x<5 ...(1)
log1/3(x−1)+log1/3(x+1)+log√3(5−x)<1
⇒−log3(x−1)−log3(x+1)+2log3(5−x)<1[∵log1/ab=−logab]
⇒log3(5−x)2(x−1)(x+1)<1
⇒(5−x)2(x−1)(x+1)<3
⇒x2−10x+25<3x2−3
⇒x2+5x−14>0
⇒x<−7 and x>2 ...(2)
From (1) & (2), we get x∈(2,5)