∣A∣=∣−2154+d(sinθ)+2(2sinθ)−d(sinθ)−2d(−sinθ)+2+2d∣
R3→R3−2R2+R1
=∣−2114+d(sinθ)+20(sinθ)−2d0∣
=1(4+d)d−(sinθ+2)(sinθ−2)
=4d+d2−sin2θ+4=(d+2)2−sin2θ
We know that 0≤sin2θ≤1
Given, the minimum value of ∣A∣=8, which is possible, when sin2θ=1
⇒(d+2)2=9
⇒d+2=±3
⇒d=1 or d=−5.