Question

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

Answer 1

Correct Answer: 55

Given \( \alpha \beta \gamma = 45 \), where \( \alpha, \beta, \gamma \) are real numbers,

The equation \( x(\alpha, 1, 2) + y(1, \beta, 2) + z(2, 3, \gamma) = (0, 0, 0) \) expands to the following system:

\[ x \alpha + y + 2z = 0, \] \[ x + y \beta + 3z = 0, \] \[ 2x + 2y + z \gamma = 0. \]

For a non-trivial solution, given \( xyz eq 0 \), the determinant of the coefficient matrix must be zero:

\[ \begin{vmatrix} \alpha & 1 & 2 \\ 1 & \beta & 2 \\ 2 & 3 & \gamma \end{vmatrix} = 0. \]

Expanding this determinant yields:

\[ \alpha \begin{vmatrix} \beta & 2 \\ 3 & \gamma \end{vmatrix} - 1 \begin{vmatrix} 1 & 2 \\ 2 & \gamma \end{vmatrix} + 2 \begin{vmatrix} 1 & \beta \\ 2 & 3 \end{vmatrix} = 0. \]

The minors are calculated as:

\[ \begin{vmatrix} \beta & 2 \\ 3 & \gamma \end{vmatrix} = \beta \gamma - 6, \] \[ \begin{vmatrix} 1 & 2 \\ 2 & \gamma \end{vmatrix} = \gamma - 6, \] \[ \begin{vmatrix} 1 & \beta \\ 2 & 3 \end{vmatrix} = 3 - 2\beta. \]

Substituting these into the determinant equation:

\[ \alpha (\beta \gamma - 6) - (\gamma - 6) + 2(3 - 2\beta) = 0. \]

Simplifying this expression:

\[ \alpha \beta \gamma - 6\alpha - \gamma + 6 + 6 - 4\beta = 0. \]

Using the given condition \( \alpha \beta \gamma = 45 \):

\[ 45 - 6\alpha - \gamma + 12 - 4\beta = 0. \]

Rearranging the terms to solve for \( \alpha, \beta, \gamma \):

\[ 6\alpha + 4\beta + \gamma = 55. \]