Let, c=ai^+bj^+ck^ Given, c×(i^+2j^+5k^)=0 ⇒i^a1j^b2k^c5=0⇒(5b−2c)i^−(5a−c)j^+(2a−b)k^=0i^+0j^+0k^ Comparing both sides, we get 5b−2c=0;5a−c=0;2a−b=0 or 5b=2c;5a=c;2a=b Also given ∣c∣2=60 ⇒a2+b2+c2=60 Putting the value of b and c in above eqn., we get a2+(2a)2+(5a)2=60⇒a2+4a2+25a2=60⇒30a2=60a2=2a=±2;b=22;c=52 Now, c=ai^+bj^+ck^ ∴c=2i^+22j^+52k^ Value of c⋅(−7i^+2j^+3k^) is (2i^+22j^+52k^)⋅(−7i^+2j^+3k^)=−72+42+152=122