The system of equations $x + y + z = 6$, $x + 2y + 3z = 10,x + 2y + \lambda z = \mu $, has no solution for

Find area of the triangle with vertices at the point given in each of the following: $(1,0),(6,0),(4,3)$

Consider the system of linear equations

$x+y+z=5, x+2 y+\lambda^2 z=9$

$x+3 y+\lambda z=\mu$, where $\lambda, \mu \in R$. Then, which of the following statement is NOT correct?

If $A = \left| {\,\begin{array}{*{20}{c}}{\sin (\theta + \alpha )}&{\cos (\theta + \alpha )}&1\\{\sin (\theta + \beta )}&{\cos (\theta + \beta )}&1\\{\sin (\theta + \gamma )}&{\cos (\theta + \gamma )}&1\end{array}\,} \right|$ ,then

If $\left| {\,\begin{array}{*{20}{c}}{{x^2} + x}&{x + 1}&{x - 2}\\{2{x^2} + 3x - 1}&{3x}&{3x - 3}\\{{x^2} + 2x + 3}&{2x - 1}&{2x - 1}\end{array}\,} \right| = Ax - 12$, then the value of $A $ is

If $\left| {\begin{array}{*{20}{c}}{x - 4}&{2x}&{2x}\\{2x}&{x - 4}&{2x}\\{2x}&{2x}&{x - 4}\end{array}} \right| = \left( {A + Bx} \right){\left( {x - A} \right)^2},$ then the ordered pair $\left( {A,B} \right) = $. . . . .