(a) Let p = a i + j + k , q = i + b j+ k and r = i + j + ck If these vectors are coplanar, then [p q r] = 0 a111b111c=0 using operation R2→R2+R1,R3→R3−R1a1−a1−a1b−1010c−1=0 Expanding along R1, a(b−1)(c−1)−(1−a)(c−1)−(1−a)(b−1)=0⇒[(a−1)+1](b−1)(c−1)+(1−a)(1−c)+(1−a)(1−b)=0⇒[(a−1)(b−1)(c−1)+(b−1)(c−1)+(1−a)(1−c)+(1−a)(1−b)=0]⇒−(1−a)(1−b)(1−c)+(−b+1)(−c+1)+(1−a)(1−c)+(1−a)(1−b)=0 On dividing both sides by (1 - a) (1 - b) (1 - c), we get −1+1−a1+1−b1+1−c1=0⇒1−a1+1−b1+1−c1=1