The area of a triangle is $13$ sq. units whose vertices are $A(8, 2)$,$B(k, 4)$,and $C(6, 7)$. Then,the integer value of $k$ is . . . . . . .

If $\alpha \neq a, \beta \neq b, \gamma \neq c$ and $\left|\begin{array}{lll}\alpha & b & c \\ a & \beta & c \\ a & b & \gamma\end{array}\right|=0$,then $\frac{a}{\alpha-a}+\frac{b}{\beta-b}+\frac{\gamma}{\gamma-c}$ is equal to :

$\left|\begin{array}{ccc}\sqrt{3} & 2 \sqrt{5} & \sqrt{5} \\ \sqrt{15} & 5 & \sqrt{10} \\ 3 & \sqrt{15} & 5\end{array}\right|=$

The set of all values of $\lambda$ for which the system of linear equations $2x_1 - 2x_2 + x_3 = \lambda x_1$,$2x_1 - 3x_2 + 2x_3 = \lambda x_2$,and $-x_1 + 2x_2 = \lambda x_3$ has a non-trivial solution:

Let ${a_2},{a_3} \in R$ such that $\left| {{a_2} - {a_3}} \right| = 6$ and $f\left( x \right) = \left| \begin{array}{ccc} 1 & {a_3} & {a_2} \\ 1 & {a_3} & {2{a_2} - x} \\ 1 & {2{a_3} - x} & {a_2} \end{array} \right|, x \in R.$ Then the greatest value of $f(x)$ is

The determinant $\left| \begin{array}{ccc} a & b & a\alpha + b \\ b & c & b\alpha + c \\ a\alpha + b & b\alpha + c & 0 \end{array} \right| = 0$,if $a, b, c$ are in