The value of $k \in R$, for which the following system of linear equations

$3 x-y+4 z=3$

$x+2 y-3 x=-2$

$6 x+5 y+k z=-3$

has infinitely many solutions, is:

If ${\Delta _1} = \left| {\,\begin{array}{*{20}{c}}x&b&b\\a&x&b\\a&a&x\end{array}\,} \right|$ and ${\Delta _2} = \left| {\,\begin{array}{*{20}{c}}x&b\\a&x\end{array}\,} \right|$ are the given determinants, then

If the system of equation $2x + 3y =\, -1; 3x + y = 2; \lambda x + 2y = \mu $ is consistent, then

Evaluate the determinant $\Delta=\left|\begin{array}{rrr}1 & 2 & 4 \\ -1 & 3 & 0 \\ 4 & 1 & 0\end{array}\right|$

If $\left| {\,\begin{array}{*{20}{c}}a&b&{a\alpha - b}\\b&c&{b\alpha - c}\\2&1&0\end{array}\,} \right| = 0$ and $\alpha \ne \frac{1}{2},$ then

If the system of equations $x - ky - z = 0$, $kx - y - z = 0$ and $x + y - z = 0$ has a non zero solution, then the possible value of k are