**Given Equations :**

2λx – 2y + 3z = 0

x + λy + 2z = 0

2x + λz = 0

**For trivial solution D = 0**

= 2λ (λ×λ – 0×2) + 2(1×λ – 2×2) + 3(1×0 – 2×λ)

= 2λ (λ^{2} – 0) + 2(λ – 4) + 3(0 – 2λ)

= 2λ 3 + 2λ – 8 – 6λ

= 2λ 3 + 4λ – 8

**Now,**

D = 0

2λ^{3} – 4λ – 8 = 0

2λ^{3} – 4λ= 8

λ(λ^{2} – 2) = 4

**Hence,**

λ = 2

**Now,**

let z = k

⇒ 4x – 2y = – 3k

And x + 2y = – 2k

**Now using the cramer’s rule**