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(D) Given the determinant equation: $\left|\begin{array}{lll}x+1 & x+2 & x+a \ x+2 & x+3 & x+b \ x+3 & x+4 & x+c\end{array}ight|=0$ Applying the r

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in $AP$

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(D) Given the determinant equation: $\left|\begin{array}{lll}x+1 & x+2 & x+a \ x+2 & x+3 & x+b \ x+3 & x+4 & x+c\end{array}ight|=0$ Applying the row operation $R_{1} ightarrow R_{1} + R_{3} - 2R_{2}$: The first row becomes: $(x+1+x+3-2(x+2), x+2+x+4-2(x+3), x+a+x+c-2(x+b))$ Simplifying the elements of the first row: $R_{1,1} = 2x + 4 - 2x - 4 = 0$ $R_{1,2} = 2x + 6 - 2x - 6 = 0$ $R_{1,3} = 2x + a + c - 2x - 2b = a + c - 2b$ The determinant becomes: $\left|\begin{array}{ccc} 0 & 0 & a+c-2b \ x+2 & x+3 & x+b \ x+3 & x+4 & x+c \end{array}ight| = 0$ Expanding along the first row: $(a+c-2b) \cdot [(x+2)(x+4) - (x+3)(x+3)] = 0$ $(a+c-2b) \cdot [x^2 + 6x + 8 - (x^2 + 6x + 9)] = 0$ $(a+c-2b) \cdot (-1) = 0$ Since $-1 
eq 0$,we must have $a+c-2b = 0$,which implies $2b = a+c$. This is the condition for $a, b, c$ to be in Arithmetic Progression $(AP)$.

If $\left|\begin{array}{lll}x+1 & x+2 & x+a \\ x+2 & x+3 & x+b \\ x+3 & x+4 & x+c\end{array}\right|=0$,then $a, b, c$ are

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