Let alpha beta gamma = 45; alpha, beta, gamma | vedclass

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(A) Given $\alpha \beta \gamma = 45$ and the system of linear equations:$x(\alpha, 1, 2) + y(1, \beta, 2) + z(2, 3, \gamma) = (0, 0, 0)$This can be wr

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$55$

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(A) Given $\alpha \beta \gamma = 45$ and the system of linear equations:$x(\alpha, 1, 2) + y(1, \beta, 2) + z(2, 3, \gamma) = (0, 0, 0)$This can be written as:$1) \alpha x + y + 2z = 0$$2) x + \beta y + 3z = 0$$3) 2x + 2y + \gamma z = 0$Since $xyz 
eq 0$,the system has a non-trivial solution,which implies the determinant of the coefficient matrix must be zero:$\begin{vmatrix} \alpha & 1 & 2 \ 1 & \beta & 3 \ 2 & 2 & \gamma \end{vmatrix} = 0$Expanding the determinant along the first row:$\alpha(\beta \gamma - 6) - 1(\gamma - 6) + 2(2 - 2\beta) = 0$$\alpha \beta \gamma - 6\alpha - \gamma + 6 + 4 - 4\beta = 0$Substitute $\alpha \beta \gamma = 45$:$45 - 6\alpha - \gamma + 10 - 4\beta = 0$$55 - (6\alpha + 4\beta + \gamma) = 0$Therefore,$6\alpha + 4\beta + \gamma = 55$.

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 $x, y, z \in R$ such that $xyz \neq 0$,then $6\alpha + 4\beta + \gamma$ is equal to:

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