=z−1=λ Any point P on this line is P(2λ,−3λ−1,λ+1) ∴ P is (2λ,3λ−1,λ+1). ∴ According to question, Distance between P(2λ,−3λ−1,λ+1)‌and(1,−1,1)‌is‌√11, ∴ √(2λ−1)2+(−3λ−1+1)2+(λ+1−1)2=√11 (2λ−1)2+(−3λ)2+(λ)2=11 4λ2−4λ+1+9λ2+λ2=11 ⇒ 14λ2−4λ−10=0⇒7λ2−2λ−5=0 ⇒ 7λ2−7λ+5λ−5=0 ⇒ 7λ(λ−1)+5(λ−1)=0 ⇒ (λ−1)(7λ+5)=0 λ=1‌or‌λ=