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(C) Given $f(x) = \left| \begin{array}{ccc} x & x+1 & x+3 \ x+2 & x+4 & x+7 \ x+6 & x+9 & x+13 \end{array} ight|$.Substitute $x = 5$ into the dete

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-$2$

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(C) Given $f(x) = \left| \begin{array}{ccc} x & x+1 & x+3 \ x+2 & x+4 & x+7 \ x+6 & x+9 & x+13 \end{array} ight|$.Substitute $x = 5$ into the determinant:$f(5) = \left| \begin{array}{ccc} 5 & 6 & 8 \ 7 & 9 & 12 \ 11 & 14 & 18 \end{array} ight|$.Expanding along the first row $(R_1)$:$f(5) = 5(9 	imes 18 - 14 	imes 12) - 6(7 	imes 18 - 11 	imes 12) + 8(7 	imes 14 - 11 	imes 9)$$f(5) = 5(162 - 168) - 6(126 - 132) + 8(98 - 99)$$f(5) = 5(-6) - 6(-6) + 8(-1)$$f(5) = -30 + 36 - 8$$f(5) = 6 - 8 = -2$.Therefore,option $(C)$ is correct.

If $f(x) = \left| \begin{array}{ccc} x & x+1 & x+3 \\ x+2 & x+4 & x+7 \\ x+6 & x+9 & x+13 \end{array} \right|$,then $f(5) =$

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