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If ∣3x47x∣=∣6321∣\begin{vmatrix} 3x & 4 \\ 7 & x \end{vmatrix} = \begin{vmatrix} 6 & 3 \\ 2 & 1 \end{vmatrix} then :

Solution

✅ Correct Option: 4

Expanding both determinants: 3x⋅x−4⋅7=6⋅1−3⋅23x \cdot x - 4 \cdot 7 = 6 \cdot 1 - 3 \cdot 2, giving 3x2−28=03x^2 - 28 = 0, so x2=283x^2 = \frac{28}{3}.

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