If area of triangle is $35$ $\mathrm{sq}$ $\mathrm{units}$ with vertices $(2,-6),(5,4)$ and $(\mathrm{k}, 4) .$ Then $\mathrm{k}$ is

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 –
 

The value of $'a'$ for which the system of equation  $a^3x + (a + 1)^3y + (a + 2)^3 z = 0$ ; $ax + (a + 1)y + (a + 2)z = 0$ ; $x + y + z = 0$  has a non-zero solution is :-

The values of the determinant $\left| {\,\begin{array}{*{20}{c}}1&{\cos (\alpha – \beta )}&{\cos \alpha }\\{\cos (\alpha – \beta )}&1&{\cos \beta }\\{\cos \alpha }&{\cos \beta }&1\end{array}\,} \right|$ is

The value of the determinant$\left| {\,\begin{array}{*{20}{c}}{ – 1}&1&1\\1&{ – 1}&1\\1&1&{ – 1}\end{array}\,} \right|$is equal to