Let $A = \left[ {\begin{array}{*{20}{c}}5&{5\alpha }&\alpha \\0&\alpha &{5\alpha }\\0&0&5\end{array}} \right]$, If ${\left| A \right|^2} = 25$, then $\left| \alpha \right|$ equals
$\frac{1}{5}$
$5$
${5^2}$
$1$
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
If $2x + 3y - 5z = 7, \,x + y + z = 6$, $3x - 4y + 2z = 1,$ then $x =$
If $x = cy + bz,\,\,y = az + cx,\,\,z = bx + ay$ (where $x, y, z $ are not all zero) have a solution other than $x = 0$, $y = 0$, $z = 0$ then $a, b$ and $ c $ are connected by the relation
If $\left| {\,\begin{array}{*{20}{c}}{{x^2} + x}&{x + 1}&{x - 2}\\{2{x^2} + 3x - 1}&{3x}&{3x - 3}\\{{x^2} + 2x + 3}&{2x - 1}&{2x - 1}\end{array}\,} \right| = Ax - 12$, then the value of $A $ is