Since, the point A(3,2,−1) is the mirror image of the point B(1,0,−1) with respect to the plane αx+βy+γz=δ, then3−1α​⇒⇒​=2−0β​=(−1)−(−1)γ​1α​γ​=1β​=0γ​=0​it is given that α+γ=1⇒α=1, so β=1.And, the mid-point of AB,M(2,1,−1) lies on the given plane, so⇒⇒∴​2α+β−γ=δ2(1)+(1)−0=δδ=3α+β=2,δ−γ=0,δ+β=4.​