$\left| {\,\begin{array}{*{20}{c}}{a - 1}&a&{bc}\\{b - 1}&b&{ca}\\{c - 1}&c&{ab}\end{array}\,} \right| = $

  • A

    $0$

  • B

    $(a - b)(b - c)(c - a)$

  • C

    ${a^3} + {b^3} + {c^3} - 3abc$

  • D

    None of these

Similar Questions

Consider the following system of questions $\alpha x+2 y+z=1$  ;  $2 \alpha x+3 y+z=1$  ;  $3 x+\alpha y+2 z=\beta$ . For some $\alpha, \beta \in R$. Then which of the following is NOT correct.

  • [JEE MAIN 2023]

If the system of linear equations $x - 2y + kz = 1$ ; $2x + y + z = 2$ ;  $3x - y - kz = 3$ Has a solution $(x, y, z) \ne 0$, then $(x, y)$ lies on the straight line whose equation is

  • [JEE MAIN 2019]

For positive numbers $x,y$ and $z$  the numerical value of the determinant $\left| {\,\begin{array}{*{20}{c}}1&{{{\log }_x}y}&{{{\log }_x}z}\\{{{\log }_y}x}&1&{{{\log }_y}z}\\{{{\log }_z}x}&{{{\log }_z}y}&1\end{array}\,} \right|$is

  • [IIT 1993]

Statement $-1$ : The system of linear equations

$x + \left( {\sin \,\alpha } \right)y + \left( {\cos \,\alpha } \right)z = 0$

$x + \left( {\cos \,\alpha } \right)y + \left( {\sin \alpha } \right)z = 0$

$x - \left( {\sin \,\alpha } \right)y - \left( {\cos \alpha } \right)z = 0$

has a non-trivial solution for only one value of $\alpha $ lying in the interval $\left( {0\,,\,\frac{\pi }{2}} \right)$ 

Statement $-2$ : The equation in $\alpha $

$\left| {\begin{array}{*{20}{c}}
  {\cos {\mkern 1mu} \alpha }&{\sin {\mkern 1mu} \alpha }&{\cos {\mkern 1mu} \alpha } \\ 
  {\sin {\mkern 1mu} \alpha }&{\cos {\mkern 1mu} \alpha }&{\sin {\mkern 1mu} \alpha } \\ 
  {\cos {\mkern 1mu} \alpha }&{ - \sin {\mkern 1mu} \alpha }&{ - \cos {\mkern 1mu} \alpha } 
\end{array}} \right| = 0$

has only one solution lying in the interval $\left( {0\,,\,\frac{\pi }{2}} \right)$

  • [JEE MAIN 2013]

If the points $(2k, k), (k, 2k)$ and $(k, k)$ with $k > 0$ enclose a triangle of area $18$ square unit then centroid of triangle is equal to