Using properties

Let

Given:

To find: the value of x using the properties of determinants

Now we will apply the operation, R₁→ R₁+R₂+R₃, we get

Taking (3a-x) common we get

Now we will apply the operation, C₁→ C₁-C₃, we get

Now we will apply the operation, C₂→ C₂-C₃, we get

Now we will expand along R₁, we get

Δ=(3a-x)[1{(0)(-2x)-(-2x)(2x)}]

Δ=(3a-x)[1(0+4x²)]

Δ=4x² (3a-x)

But Given Δ=0

So,

4x² (3a-x)=0

⇒ 4x²=0 or (3a-x)=0

⇒x=0 or x=3a

Hence this is the required values of x