If $\left| {\,\begin{array}{*{20}{c}}{{a_1}}&{{b_1}}&{{c_1}}\\{{a_2}}&{{b_2}}&{{c_2}}\\{{a_3}}&{{b_3}}&{{c_3}}\end{array}\,} \right| = 5$; then the value of $\left| {\,\begin{array}{*{20}{c}}{{b_2}{c_3} - {b_3}{c_2}}&{{c_2}{a_3} - {c_3}{a_2}}&{{a_2}{b_3} - {a_3}{b_2}}\\{{b_3}{c_1} - {b_1}{c_3}}&{{c_3}{a_1} - {c_1}{a_3}}&{{a_3}{b_1} - {a_1}{b_3}}\\{{b_1}{c_2} - {b_2}{c_1}}&{{c_1}{a_2} - {c_2}{a_1}}&{{a_1}{b_2} - {a_2}{b_1}}\end{array}\,} \right|$is

If the system of equations

$x+y+z=2$

$2 x+4 y-z=6$

$3 x+2 y+\lambda z=\mu$ has infinitely many solutions, then 

Let the area of the triangle with vertices $A (1, \alpha)$, $B (\alpha, 0)$ and $C (0, \alpha)$ be $4\, sq.$ units. If the point $(\alpha,-\alpha),(-\alpha, \alpha)$ and $\left(\alpha^{2}, \beta\right)$ are collinear, then $\beta$ is equal to

The system of equations $kx + 2y\,-z = 1$  ;  $(k\,-\,1)y\,-2z = 2$  ;  $(k + 2)z = 3$ has unique solution, if $k$ is equal to

The value of a for which the system of equations ${a^3}x + {(a + 1)^3}y + {(a + 2)^3}z = 0,$ $ax + (a + 1)y + (a + 2)z = 0,$ $x + y + z = 0,$ has a non zero solution is

Which of the following is correct?