The correct option is C 0,±√32(a2+b2+c2)
∣∣
∣∣a−xcbcb−xabac−x∣∣
∣∣=0⇒∣∣
∣∣a+b+c−xcba+b+c−xb−xaa+b+c−xac−x∣∣
∣∣=0⇒x=∑a=0 (by hypothesis)
or 1{(b−x)(c−x)−a2}−c{c−x−a}+b{a−b+x}=0 by expanding the determinant.
or x2−(a2+b2+c2)+(ab+bc+ca)=0
or x2−(∑a2)−12(∑a2)=0{∵a+b+c=0⇒(a+b+c)2=0⇒∑a2+2∑ab=0∑ab=−12∑a2}or x=±√32∑a2
∴ The solution is x= 0 or ±√32∑a2
Trick : Put a=1, b=-1 and c=0 so that they satisfy the condition a+b+c=0. Now the determinant becomes ∣∣
∣∣1−x0−10−1−x1−11−x∣∣
∣∣=0⇒(1−x){x(1+x)−1}+1(1+x)=0⇒(1−x){x2+x−1}+x+1=0⇒x(x2−3)=0
Now putting these in the options, we find that option (c) gives the same values i.e., 0, ±√3.