Option (b) is correct

Area of given triangle = \(\frac{1}{2}\)\(\begin{vmatrix} 2 & -2 & 1 \\ 7 & 10 & 1 \\ 4 & 6 & 1 \end{vmatrix}\) 

= \(\frac{1}{2}\) \(\big(\)2(10 - 6) + 2(7 - 4) + 1(42 - 40)\(\big)\)

= \(\frac{1}{2}\)(8 + 6 + 2) = \(\frac{16}{2}\) = 8

Area of triangle in option A =  \(\frac{1}{2}\)\(\begin{vmatrix} -4 & 2 & 1 \\ 8 & 7 & 1 \\ 1 & 1 & 1 \end{vmatrix}\)

= \(\frac{1}{2}\) \(\big(\)-4(7 - 1) - 2(8 - 1) + 1(8 - 7)\(\big)\)

= \(\frac{1}{2}\)(-24 - 14 + 1)

= \(\frac{1}{2}\)(-37) = \(\frac{37}{2}\) \(\ne\) 8 (Area always positive)

Hence, not congruent to given triangle

Area of triangle given in option B = \(\frac{1}{2}\) \(\begin{vmatrix} -5 & 3 & 1 \\ 7 & -2 & 1 \\ 3 & 1 & 1 \end{vmatrix}\)

= \(\frac{1}{2}\)\(\big(\)-5(-2 - 1) - 3(7 - 3) + 1(7 + 6)\(\big)\)

= \(\frac{1}{2}\) (15 - 12 + 13)

= \(\frac{1}{2}\) x 16 = 8 (It my be congruent to given triangle)

Area of triangle given in option (c) = \(\frac{1}{2}\)\(\begin{vmatrix} 9 & 6 & 1 \\ 1 & 8 & 1 \\ -1 & 4 & 1 \end{vmatrix}\)

= \(\frac{1}{2}\)\(\big(\)9(8 - 4) - 6(1 + 1) + 1(4 + 8)\(\big)\)

= \(\frac{1}{2}\)(36 - 12 +12)

= \(\frac{36}{2}\) = 18 \(\ne\)8

Hence not congruent to given triangle

Area of triangle given in option (d) = \(\frac{1}{2}\)\(\begin{vmatrix} -7 & -5 & 1 \\ -3 & -2 & 1 \\ 3 & 3 & 1 \end{vmatrix}\)

= \(\frac{1}{2}\)\(\big(\)-7(-2 - 3) + 5(-3 - 3) + 1(-9 + 6)\(\big)\)

= \(\frac{1}{2}\)(+35 - 45 - 3)

= \(\frac{1}{2}\) x 13 \(\ne\) 8 (Area always positive)

Hence, not congruent to given triangle

Since, only area of triangle given in option (b) is equal to Area of given triangle.

Therefore, only possibility for congruence.