Solution:

Given expression: x/(bc-a) = y/(ca-b) = z/(ab-c)

Let's assume, x/(bc-a) = y/(ca-b) = z/(ab-c) = k (some constant)

Therefore, x = k(bc-a), y = k(ca-b), and z = k(ab-c)

Now, let's simplify the given expression: (bc)x - (c-a)y + (a-b)z

= (bc)(k(bc-a)) - (c-a)(k(ca-b)) + (a-b)(k(ab-c))

= k[(bc)^2 - a(bc) + (ca - b)(c - a) + (ab - c)(b - a)]

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

= k[(bc)^2 - a(bc) + ac^2 - abc - acc + aab + abc - ab^2 - aac + ab - aab + abc - b^2c + ac - bc + c^2 - ab + bc - c^2]

= k[(bc)^2 - a(bc) - b^2c + ac - ab]

= k[bc(bc - a) - b^2c + ac - ab]

= k[bc(bc - a - b) + ac - ab]

= k[bc(-a - b + c) + ac - ab]

= k[(c - a)bc + ac - ab]

= k[(c - a)(bc - a) + a(c - b)]

= k[(c - a)x + ay]

= k[(c - a)(k(bc - a)) + k(ca - b)]

= k^2[(c - a)(bc - a) + (ca - b)]

= k^2[(c - a)(bc - a) + (c - a)(a - b)]

= k^2(c - a)(bc - a + a - b)

= k^2(c - a)(bc - b)

Now, we know that x/(bc-a) = y/(ca-b) = z/(ab-c) = k

Therefore, x + y + z = k(bc-a) + k(ca-b) + k(ab-c)

= k(bc + ca + ab - a - b - c)

= k[(b + c)a - bc - a - b - c]

= k[(b + c - 1)a - (b + c)]

= k[(b + c - 1)(bc - b)]

Therefore, (bc)x - (c-a)y + (a-b)z = k^2(c - a)(bc - b)

and x + y + z = k[(b + c - 1)(bc - b)]

Since k is a constant, (bc)x - (c-a)y + (a-b)z and x + y + z will also be constant.

Hence, the answer is (b) 0.