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(B) For a homogeneous system of linear equations $AX = 0$ to have a non-trivial solution, the determinant of the coefficient matrix $A$ must be zero,

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$6$

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(B) For a homogeneous system of linear equations $AX = 0$ to have a non-trivial solution, the determinant of the coefficient matrix $A$ must be zero, i.e.,$|A| = 0$.The coefficient matrix is given by:$A = \begin{bmatrix} 1 & -2 & 3 \ 2 & 4 & -5 \ 3 & \lambda & \mu \end{bmatrix}$Setting the determinant to zero:$\begin{vmatrix} 1 & -2 & 3 \ 2 & 4 & -5 \ 3 & \lambda & \mu \end{vmatrix} = 0$Expanding along the first row:$1(4\mu - (-5\lambda)) - (-2)(2\mu - (-15)) + 3(2\lambda - 12) = 0$$1(4\mu + 5\lambda) + 2(2\mu + 15) + 3(2\lambda - 12) = 0$$4\mu + 5\lambda + 4\mu + 30 + 6\lambda - 36 = 0$$(4\mu + 4\mu) + (5\lambda + 6\lambda) + (30 - 36) = 0$$8\mu + 11\lambda - 6 = 0$$8\mu + 11\lambda = 6$

If the homogeneous system of linear equations $x-2y+3z=0, 2x+4y-5z=0, 3x+\lambda y+\mu z=0$ has a non-trivial solution, then $8\mu+11\lambda=$

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