Correct option is (A) 0
\(x = 2p-3q,\)
\(y=p+2q\)
and \(z = q - 3p\)
\(\therefore\) x+y+z = (2p - 3q) + (p + 2q) + (q - 3p)
= (2p + p - 3p) + (-3q + 2q + q)
= (3p - 3p) + (-3q + 3q)
= 0+0 = 0
Now, \(x^2+y^2+z^2+2yz+2zx+2xy\) \(=(x+y+z)^2\)
\(=0^2=0\)