The correct option is A {−1}
|x−|3−x||−3x=8
Case−1:
x>3
⇒|x−(−(3−x))|−3x=8
⇒|x+3−x|−3x=8
⇒x=−53
But x shall be >3
So, x∈ϕ
Case−2:
x≤3
⇒|x−(3−x)|−3x=8
⇒|2x−3|−3x=6
Case−2a:
x<32
⇒x∈(−∞,32)
⇒−(2x−3)−3x=8
⇒−5x=5
⇒x=−1∈(−∞,32)
Case−2b:
x≥32
x∈[32,3]
⇒(2x−3)−3x=8
⇒−x=11
⇒x=−11∉[32,3]
Final solution = case 1 ∪ case 2
=ϕ∪{−1}={−1}