QJEEM24Ma10de: JEE Main 2024 Mathematics - Class 12 JEEM24Ma10de Solution: Let . If for some , then is equal to _________.

Given that $\alpha \beta \gamma = 45$ and $\alpha, \beta, \gamma \in \mathbb{R}$ , consider the equation $x(\alpha, 1, 2) + y(1, \beta, 2) + z(2, 3, \gamma) = (0, 0, 0)$ for some $x, y, z \in \mathbb{R}$ where $x y z \neq 0$ . To find the value of $6 \alpha + 4 \beta + \gamma$ , follow these steps:

1. Express the given equation in matrix form:

$\begin{aligned} & \alpha x + y + 2 z = 0 \\ & x + \beta y + 3 z = 0 \\ & 2 x + 2 y + \gamma z = 0 \end{aligned}$

1. Since $x, y, z \neq 0$ , the determinant of the coefficients matrix must be zero:

$\left|\begin{array}{ccc} \alpha & 1 & 2 \\ 1 & \beta & 3 \\ 2 & 2 & \gamma \end{array}\right| = 0$

1. Calculate the determinant of the matrix:

$\alpha \beta \gamma - 6 \alpha - 4 \beta - \gamma + 10 = 0$

1. Given $\alpha \beta \gamma = 45$ , substitute this value into the equation:

$45 - 6 \alpha - 4 \beta - \gamma + 10 = 0$

1. Simplify the equation:

$45 + 10 = 6 \alpha + 4 \beta + \gamma$

1. Thus,

$6 \alpha + 4 \beta + \gamma = 55$

So, the value of $6 \alpha + 4 \beta + \gamma$ is 55.