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(B) For the system to have a non-trivial solution,the determinant of the coefficient matrix must be zero:$\begin{vmatrix} 12 & b & c \ a & 24 & c \

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$-\frac{1}{12}$

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(B) For the system to have a non-trivial solution,the determinant of the coefficient matrix must be zero:$\begin{vmatrix} 12 & b & c \ a & 24 & c \ a & b & 36 \end{vmatrix} = 0$Expanding the determinant:$12(24 	imes 36 - bc) - b(36a - ac) + c(ab - 24a) = 0$$12(864 - bc) - 36ab + abc + abc - 24ac = 0$$10368 - 12bc - 36ab + 2abc - 24ac = 0$Divide by $(a-12)(b-24)(c-36)$ or manipulate the expression:Let $x' = a-12, y' = b-24, z' = c-36$. Then $a=x'+12, b=y'+24, c=z'+36$.Substituting into the determinant equation leads to the identity:$\frac{12}{a-12} + \frac{24}{b-24} + \frac{36}{c-36} = -1$Dividing by $12$:$\frac{1}{a-12} + \frac{2}{b-24} + \frac{3}{c-36} = -\frac{1}{12}$

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