Concept:Apply the distributive property to remove brackets, then combine like terms.Explanation:Expand each term using a(b+c)=ab+ac.First, 3(5a2+2c)=3×5a2+3×2c=15a2+6c.Next, expand −2a(1−3a). Since −×−=+, we get:−2a(1−3a)=−2a×1+(−2a)(−3a)=−2a+6a2.Now substitute into the expression:3(5a2+2c)−2a(1−3a)−6c=15a2+6c−2a+6a2−6c.Group like terms:=(15a2+6a2)−2a+(6c−6c)=21a2−2a+0.The c terms cancel, leaving 21a2−2a.Answer:21a2−2aTherefore, the correct option is D.