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3 and 4 .Determinants and Matrices
hard
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..............
A
$55$
B
$56$
C
$54$
D
$31$
(JEE MAIN-2024)
Solution
$ \alpha \beta \gamma=45, \alpha \beta \gamma \in R $
$ x(\alpha, 1,2)+y(1, \beta, 2)+z(2,3, \gamma)=(0,0,0) $
$ x, y, z \in R, x y z \neq 0 $
$ \alpha x+y+2 z=0 $
$ x+\beta y+3 z=0 $
$ 2 x+2 y+\gamma z=0 $
$ x y z \neq 0 \Rightarrow$ non-trivial
$\left|\begin{array}{lll}\alpha & 1 & 2 \\ 1 & \beta & 3 \\ 2 & 2 & \gamma\end{array}\right|=0$
$ \Rightarrow \alpha(\beta \gamma-6)-1(\gamma-6)+2(2-2 \beta)=0 $
$ \Rightarrow \alpha \beta \gamma-6 \alpha-\gamma+6+4-4 \beta=0 $
$ \Rightarrow 6 \alpha+4 \beta+\gamma=55$
Standard 12
Mathematics
