If $a + b + c = 0$, then the solution of the equation $\left| {\,\begin{array}{*{20}{c}}{a - x}&c&b\\c&{b - x}&a\\b&a&{c - x}\end{array}\,} \right| = 0$ is

  • A

    $0$

  • B

    $ \pm \frac{3}{2}({a^2} + {b^2} + {c^2})$

  • C

    $0,\, \pm \sqrt {\frac{3}{2}({a^2} + {b^2} + {c^2})} $

  • D

    $0,\,\, \pm \sqrt {{a^2} + {b^2} + {c^2}} $

Similar Questions

By using properties of determinants, show that:

$\left|\begin{array}{lll}x & x^{2} & y z \\ y & y^{2} & z x \\ z & z^{2} & x y\end{array}\right|=(x-y)(y-z)(z-x)(x y+y z+z x)$

Verify Property $2$ for $\Delta=\left|\begin{array}{ccc}2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7\end{array}\right|$

If the minimum and the maximum values of the function $f :\left[\frac{\pi}{4}, \frac{\pi}{2}\right] \rightarrow R ,$ defined by : 

$f (\theta)=\left|\begin{array}{ccc}-\sin ^{2} \theta & -1-\sin ^{2} \theta & 1 \\ -\cos ^{2} \theta & -1-\cos ^{2} \theta & 1 \\ 12 & 10 & -2\end{array}\right|$ are $m$ and $M$ respectively, then the ordered pair $( m , M )$ is equal to

  • [JEE MAIN 2020]

$$f(x)=\left| {\begin{array}{*{20}{c}} {{{\sin }^2}x}&{ - 2 + {{\cos }^2}x}&{\cos 2x} \\ {2 + {{\sin }^2}x}&{{{\cos }^2}x}&{\cos 2x} \\ {{{\sin }^2}x}&{{{\cos }^2}x}&{1 + \cos 2x} \end{array}} \right| ,x \in[0, \pi]$$

Then the maximum value of $f(x)$ is equal to $.....$

  • [JEE MAIN 2021]

By using properties of determinants, show that:

$\left|\begin{array}{ccc}
1 & 1 & 1 \\
a & b & c \\
a^{3} & b^{3} & c^{3}
\end{array}\right|=(a-b)(b-c)(c-a)(a+b+c)$