QJEEM20Maa8f6: JEE Main 2020 Mathematics - Class 12 JEEM20Maa8f6 Solution: If the system of linear equations, x + y + z = 6 x + 2y + 3z =…

Given system of equation more than 2 solutions. Hence system of equation has infinite many solution.

$$\therefore$$ $$\mathrm{\Delta }$$ = $$\mathrm{\Delta }$$₁ = $$\mathrm{\Delta }$$₂ = $$\mathrm{\Delta }$$₃ = 0

$$\mathrm{\Delta }$$ = $$\left|\begin{array}{ccc}1& 1& 1\\ 1& 2& 3\\ 3& 2& \lambda \end{array}\right|$$ = 0

$$\Rightarrow$$ 1(2λ – 6) – 1(λ – 9) + 1(– 4) = 0

$$\Rightarrow$$ 2λ – 6 – λ + 9 – 4 = 0

$$\Rightarrow$$ λ = 1

$$\mathrm{\Delta }$$₁ = $$\left|\begin{array}{ccc}6& 1& 1\\ 10& 2& 3\\ \mu & 2& \lambda \end{array}\right|$$ = 0
6(2λ – 6) – 1(10λ – 3μ) + 1(20 – 2μ) = 0

$$\Rightarrow$$ 12λ – 36 – 10λ + 3μ + 20 – 2μ = 0

$$\Rightarrow$$ 2λ + μ = 16

$$\Rightarrow$$ 2 + μ = 16

$$\Rightarrow$$ $$\mu$$ = 14

$$\therefore$$ $$\mu$$ - $$\lambda$$² = 14 - 1 = 13