If $\frac{x}{(b+c)(b+c-2a)}=\frac{y}{(c-a) (c + a - 2b)}=\frac{z}{(a - b)(a + b - 2c)}$ then value of $x+y+z$ is:

0

Let,

$\frac{x}{(b+c)(b+c-2a)}=\frac{y}{(c-a) (c + a - 2b)}=\frac{z}{(a - b)(a + b - 2c)}$ = k

Re arranging,
x= k (b-c)(b+c-2a)
y= k (c-a)(c+a-2b)
z= k (a-b)(a+b-2c)

now x+y+z = k [ (b-c)(b+c-2a) + (c-a)(c+a-2b) + (a-b)(a+b-2c) ]

so x+y+z= k[ b*b+ bc-2ab-bc-c*c+2ac+c*c+ac-2bc-ac-a.a+2ab+ a*a +a*b-2ac-ab-b*b+2bc]

x+y+z = k * 0

The correct answer is Option (3) → 0