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If ∣x−22x−33x−4x−42x−93x−16x−82x−273x−64∣=0\begin{vmatrix}x-2 & 2x-3 & 3x-4\\x-4 & 2x-9 & 3x-16\\x-8 & 2x-27 & 3x-64\end{vmatrix} = 0 then the value of x is:

Solution

✅ Correct Option: 4

Apply R2→R2−R1R_2 \to R_2 - R_1 and R3→R3−R1R_3 \to R_3 - R_1 to get rows (−2,−6,−12)(-2,-6,-12) and (−6,−24,−60)(-6,-24,-60), then R3→R3−3R2R_3 \to R_3 - 3R_2 gives (0,−6,−24)(0,-6,-24).

Expanding along this row yields 12x−48=012x - 48 = 0, so x=4x = 4.

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