`a ^(2) + bc + ab + ac=` ?

Find the square root of : 4 a ^(2) + 9b ^(2) + c ^(2) - 12 ab + 6 bc - 4 ac

| [-bc, b ^ (2) + bc, c ^ (2) + bca ^ (2) + ac, -ac, c ^ (2) + aca ^ (2) + ab, b ^ (2) + ab, -ab (ab + bc + ac), is = 64. then

If a = 24, b = 26, c = 28 , then the value of a^2 + b^2 + c^2 - ab - bc - ac will be

If ∣ -a a^2 ab ac ab -b^2 bc ac bc -c^2 | = ka^2b^2c^2 , then k is equal to

Subtract : 2 ab + 5 bc - 7 ac from 5 ab - 2 bc - 2 ac + 10 abc

If a, b and c are in H.P., then the value of ((ac + ab - bc)(ab + bc - ac))/((abc)^(2)) is -

Prove |[-bc, b^2+bc, c^2+bc] , [a^2+ac, -ac, c^2+ac] , [a^2+ab, b^2+ab, -ab]|=(ab+bc+ca)^2

The determinant |(a^(2)+10,ab, ac),(ab, b^(2)+10, bc),(ac, bc, c^(2)+10)| is

The determinant Delta = |(a^(2) + x^(2),ab,ac),(ab,b^(2) + x^(2),bc),(ac,bc,c^(2) + x^(2))| is divisible