The values of $a$ and $b$, for which the system of equations $2 x+3 y+6 z=8$ ; $x+2 y+a z=5$ ; $3 x+5 y+9 z=b$ has no solution, are:
$a=3, b=13$
$a \neq 3, b \neq 13$
$a \neq 3, b=3$
$a=3, b \neq 13$
$\left| {\,\begin{array}{*{20}{c}}{{a_1}}&{m{a_1}}&{{b_1}}\\{{a_2}}&{m{a_2}}&{{b_2}}\\{{a_3}}&{m{a_3}}&{{b_3}}\end{array}\,} \right| = $
If $\left| {\begin{array}{*{20}{c}}{a\, + \,1}&{a\, + \,2}&{a\, + \,p}\\{a\, + \,2}&{a\, +\,3}&{a\, + \,q}\\{a\, + \,3}&{a\, + \,4}&{a\, + \,r}\end{array}} \right|$ $= 0$ , then $p, q, r$ are in :
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 $a \ne 6,b,c$ satisfy $\left| {\,\begin{array}{*{20}{c}}a&{2b}&{2c}\\3&b&c\\4&a&b\end{array}\,} \right| = 0,$then $abc = $
The existence of unique solution of the system of equations, $x+y+z=\beta $ , $5x-y+\alpha z=10$ , $2x+3y-z=6$ depends on