Let $\alpha \beta \gamma=45 ; \alpha, \beta, \gamma \in R$. If $x(\alpha, 1,2)+y(1, \beta, 2)$ $+z(2,3, \gamma)=(0,0,0)$ for some $x, y, z \in R, x y z \neq$ 0 , then $6 \alpha+4 \beta+\gamma$ is equal to..............

If ${\Delta _1} = \left| {\,\begin{array}{*{20}{c}}x&b&b\\a&x&b\\a&a&x\end{array}\,} \right|$ and ${\Delta _2} = \left| {\,\begin{array}{*{20}{c}}x&b\\a&x\end{array}\,} \right|$ are the given determinants, then

For the system of linear equations

$2 x-y+3 z=5$

$3 x+2 y-z=7$

$4 x+5 y+\alpha z=\beta$

Which of the following is NOT correct ?

If the system of linear equations $2 x+3 y-z=-2$  ; $x+y+z=4$  ; $x-y+|\lambda| z=4 \lambda-4$  (where $\lambda \in R$), has no solution, then

Find area of the triangle with vertices at the point given in each of the following: $(-2,-3),(3,2),(-1,-8)$

The determinant $\,\left| {\,\begin{array}{*{20}{c}}1&1&1\\1&2&3\\1&3&6\end{array}\,} \right|$ is not equal to