$\left| {\,\begin{array}{*{20}{c}}{a - 1}&a&{bc}\\{b - 1}&b&{ca}\\{c - 1}&c&{ab}\end{array}\,} \right| = $
$0$
$(a - b)(b - c)(c - a)$
${a^3} + {b^3} + {c^3} - 3abc$
None of these
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.
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
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
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 $\left( {0\,,\,\frac{\pi }{2}} \right)$
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