Equation of plane P₃ will be (x + z – 1) + λy = 0

⇒ x + λy + z – 1 = 0

Distance from (0, 1, 0) is equal to 1

$\Rightarrow \frac{|0+\mathrm{\lambda }\mathrm{·}1+0-1|}{\sqrt{1+{\mathrm{\lambda }}^2+1}}=\text{ 1}$

⇒ (λ – 1)² = λ² + 2 ⇒ λ² – 2λ + 1 = λ² + 2

$\Rightarrow \text{2}\mathrm{\lambda }=–\text{ 1}\Rightarrow \mathrm{\lambda }=\frac{-1}{2}.$

$\therefore$  Plane will be x – $\frac{\mathrm{y}}{2}$ + z – 1 = 0 ⇒ 2x – y + 2z – 2 = 0

Distance from (α, β, γ) = 2

⇒ $\frac{|2\mathrm{\alpha }-\mathrm{\beta }+2\mathrm{\gamma }-2|}{\sqrt{2^2+1^2+2^2}}$ = 2 ⇒ 2α – β + 2γ – 2 = ± 6

⇒ 2α – β + 2γ = 8          or         2α – β + 2γ = – 4.