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If ∣2a2a3b+c3b2a+c3b2a+3bcc∣=48\begin{vmatrix}2a & 2a & 3b+c\\3b & 2a+c & 3b\\2a+3b & c & c\end{vmatrix} = 48, then abc is equal to

Solution

✅ Correct Option: 4

Put p=2ap=2a, q=3bq=3b, r=cr=c. Expanding the determinant of [ppq+rqp+rqp+qrr]\begin{bmatrix}p&p&q+r\\q&p+r&q\\p+q&r&r\end{bmatrix} gives −4pqr-4pqr. So −4(2a)(3b)(c)=−24abc=48-4(2a)(3b)(c)=-24abc=48, giving abc=−2abc=-2.

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