Let the system of linear equations  $x+y+k z=2$ ;  $2 x+3 y-z=1$ ; $3 x+4 y+2 z=k$ , have infinitely many solutions. Then the system $( k +1) x +(2 k -1) y =7$ ; $(2 k +1) x +( k +5) y =10 \text { has : }$

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

If $\alpha , \beta \, and \, \gamma$ are real numbers , then $D = \left|{\begin{array}{*{20}{c}}1&{\cos \,(\beta \, – \,\alpha )}&{\cos \,(\gamma \, – \,\alpha )}\\{\cos \,(\alpha \, – \,\beta )}&1&{\cos \,(\gamma \, – \,\beta )}\\{\cos \,(\alpha \, – \,\gamma )}&{\cos \,(\beta \, – \,\gamma )}&1 \end{array}} \right|$ =

If $a_i^2 + b_i^2 + c_i^2 = 1,\,i = 1,2,3$ and $a_ia_j + b_ib_j +c_ic_j = 0$ $\left( {i \ne j,i,j = 1,2,3} \right)$ then the value of determinant $\left| {\begin{array}{*{20}{c}}
  {{a_1}}&{{a_2}}&{{a_3}} \\ 
  {{b_1}}&{{b_2}}&{{b_3}} \\ 
  {{c_1}}&{{c_2}}&{{c_3}} 
\end{array}} \right|$ is