Using properties of determinants solve the following for x:

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

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


Taking (3x+a) 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=(3x+a)[1{(0)(-a)-(-a)(a)}]
Δ=(3x+a)[1(0+a2)]
Δ=a2 (3x+a)
Δ=a2 (3x+a)
But Given Δ=0
So,
a2 (3x+a)=0
⇒ (3x+a)=0
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Hence this is the required values of x
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