We know that,
Domain of sec−1x is (−∞,−1]∪[1,∞).
Now, f(x)=sec−1(5x+32x)
So,
5x+32x≤−1or5x+32x≥1
⇒5x+32x+1≤0or5x+32x−1≥0
⇒5x+37x+3≤0or5x+3−3x−3≥0
⇒5x+37x+3≤0or5x+3x+1≤0
⇒x∈[−1,−53)∪(−53,−73]
So,
α=−1,β=−53,γ=−53,δ=−73
Now,
∣3α+10(β+γ)+21δ∣=∣−3−6−6−9∣=24
Hence this is the required answer.