$\left| {\,\begin{array}{*{20}{c}}1&1&1\\1&{1 + x}&1\\1&1&{1 + y}\end{array}\,} \right| = $
Suppose $D = \left| {\,\begin{array}{*{20}{c}}{{a_1}}&{{b_1}}&{{c_1}}\\{{a_2}}&{{b_2}}&{{c_2}}\\{{a_3}}&{{b_3}}&{{c_3}}\end{array}\,} \right|$ and $D' = \left| {\,\begin{array}{*{20}{c}}{{a_1} + p{b_1}}&{{b_1} + q{c_1}}&{{c_1} + r{a_1}}\\{{a_2} + p{b_2}}&{{b_2} + q{c_2}}&{{c_2} + r{a_2}}\\{{a_3} + p{b_3}}&{{b_3} + q{c_3}}&{{c_3} + r{a_3}}\end{array}\,} \right|$, then
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
If system of equations $kx + 2y - z = 2,$$\left( {k - 1} \right)x + ky + z = 1,x + \left( {k - 1} \right)y + kz = 3$ has only one solution, then number of possible real value$(s)$ of $k$ is -
$\left| {\,\begin{array}{*{20}{c}}x&4&{y + z}\\y&4&{z + x}\\z&4&{x + y}\end{array}\,} \right| = $