If the reflection of the point P(1,0,0) in the line x−12=y+1−3=z+108 is (α,β,γ), then α+β+γ is

Given reflection of the point P(1,0,0) in the line x−12=y+1−3=z+108 is Q(α,β,γ) Direction ratios of PQ¯ is α−1,β−0,γ−0=α−1,β,γDirection ratios of the given line is 2,−3,8 Perpendicular to the given line PQ¯⇒2α−1+−3β+8γ=0 ( ∵ Two lines are perpendicular whose direction ratios are a1,b1,c1 and a2,b2,c2 then a1a2+b1b2+c1c2=0)⇒2α−3β+8γ=2……………1 Midpoint of PQ¯ is α+12,β2,γ2 lies on the given line ⇒α+12−12=β2+1−3=γ2+108=λ⇒α−14=β+2−6=γ+2016=λ⇒α=4λ+1,β=−6λ−2,γ=16λ−20 From (1)⇒2(4λ+1)−3(−6λ−2)+8(16λ−20)=2⇒154λ=154⇒λ=1∴α=5,β=−8,γ=−4α+β+γ=−7