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If ∣x3xx2x7xx7∣=ax4+bx3+cx2+dx+e\begin{vmatrix} x & 3 & x \\ x^2 & x & 7 \\ x & x & 7 \end{vmatrix} = ax^4+bx^3+cx^2+dx+e, then the value of 6a+5b+4c+3d+2e6a+5b+4c+3d+2e is

Solution

✅ Correct Option: 4

Expanding along the first row:

x(7x−7x)−3(7x2−7x)+x(x3−x2)x(7x-7x) - 3(7x^2-7x) + x(x^3-x^2)

=0−21x2+21x+x4−x3= 0 - 21x^2 + 21x + x^4 - x^3.

So a=1a=1, b=−1b=-1, c=−21c=-21, d=21d=21, e=0e=0.

Therefore 6a+5b+4c+3d+2e=6−5−84+63+0=−206a+5b+4c+3d+2e = 6 - 5 - 84 + 63 + 0 = -20.

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