The value of $\lambda$ and $\mu$ such that the system of equations $x+y+z=6,3 x+5 y+5 z=26, x+2 y+\lambda z=\mu$ has no solution, are :

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

If $f(\theta ) =\left| {\begin{array}{*{20}{c}}
1&{\cos {\mkern 1mu} \theta }&1\\
{ - \sin {\mkern 1mu} \theta }&1&{ - \cos {\mkern 1mu} \theta }\\
{ - 1}&{\sin {\mkern 1mu} \theta }&1
\end{array}} \right|$ and $A$ and $B$ are respectively the maximum and the minimum values of $f(\theta )$, then $(A , B)$ is equal to

If $|A|$  denotes the value of the determinant of the square matrix $A$ of order $3$ , then $ |-2A|=$

If $A=\left[\begin{array}{lll}1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9\end{array}\right],$ find $|A|$.

The area of a triangle is $5$ and two of its vertices are $A(2, 1), B(3, -2)$. The third  vertex which lies on line $y = x + 3$ is-