If […] =0 , then x is equal to

The correct answer is Option (2) → \( \pm\displaystyle\frac{1}{\sqrt{2}}\sqrt{(a-b)^2+(b-c)^2+(c-a)^2} \)

With \( C_2\to C_2-C_1 \) and \( C_3\to C_3-C_1 \), the equation becomes

    \( \displaystyle \begin{vmatrix}1 & 0 & 0\\c & b-c-x & a-c\\b & a-b & c-b-x\end{vmatrix}=0 \)

    \( \displaystyle (b-c-x)(c-b-x)-(a-c)(a-b)=0 \)

The first product is \( x^2-(b-c)^2 \), so

    \( \displaystyle x^2=(b-c)^2+(a-b)(a-c) \)

    \( \displaystyle x^2=a^2+b^2+c^2-ab-bc-ca \)

Using the identity

    \( \displaystyle a^2+b^2+c^2-ab-bc-ca=\frac{1}{2}\left[(a-b)^2+(b-c)^2+(c-a)^2\right] \)

we get

    \( \displaystyle x=\pm\frac{1}{\sqrt{2}}\sqrt{(a-b)^2+(b-c)^2+(c-a)^2} \)