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If $ab + bc + ca = 0$ and $\left| {\,\begin{array}{*{20}{c}}{a - x}&c&b\\c&{b - x}&a\\b&a&{c - x}\end{array}\,} \right| = 0$, then one of the value of $x$ is
${({a^2} + {b^2} + {c^2})^{\frac{1}{2}}}$
${\left[ {\frac{3}{2}({a^2} + {b^2} + {c^2})} \right]^{\frac{1}{2}}}$
${\left[ {\frac{1}{2}({a^2} + {b^2} + {c^2})} \right]^{\frac{1}{2}}}$
None of these
Solution
(a) Applying $ {c_1}→{c_1}+{c_2}+{c_3}$
$(a+b+c-x)$ $\left| {\,\begin{array}{*{20}{c}}
1&c&b \\
1&{b – x}&a \\
1&a&{c – x}
\end{array}\,} \right|\, = 0$
=> $(a+b+c-x)$ $[{(b-x)(c-x)-a^2}$ $ + c(a-c+x) + {b({a-b+x)}}] =0$
=> $(a+b+c-x)$ $[(bc-cx+bx+x^2-a^2 + ca – c^2 + cx +ab -b^2 +bx] =0$
==> $(a+b+c)[x^2 -(a^2+b^2+c^2)+ab+bc+ca]=0$
==>$(a+b+c-x)[x^2-(a^2 + b^2 +c^2] = 0$
[ $\because$ $ab + bc + ca = 0$]
$\therefore$ $x=a+b+c $ $ and (a^2+b^2+c^2)^{1/2} $
