If $$\frac{y}{x+z} = \frac{x-y}{z} = \frac{z}{x} \quad ; \quad x,y,z \neq 0$$​​

then​$$\frac{1}{x} : \frac{1}{y} : \frac{1}{z}$$​ is equal to ?

Correct option is D

Given :

$$\frac{y}{x+z} = \frac{x-y}{z} = \frac{z}{x} = k \ (\text{say})$$​​

Solution :

If $$\frac{a}{b} = \frac{c}{d} = k$$​​

a = bk, c = dk

From
​$$\frac{z}{x} = k$$​ 

z = kx .....(1)

From
​$$\frac{y}{x+z} = k$$​

y = k(x+z)

Substitute z = kx:
​$$y = k(x + kx) = kx(1+k) \quad (2)$$​​

From
​$$\frac{x-y}{z} = k$$​

x - y = kz

Substitute (1) and (2):
x - kx(1+k) = k(kx)

​$$x(1 - k - k^2) = k^2x$$​​

​$$1 - k - k^2 = k^2$$​​

​$$2k^2 + k - 1 = 0$$​​

(2k - 1)(k + 1) = 0
k =$$\frac{1}{2}$$​​

Now find (x : y : z)

From (1):
z =$$\frac{x}{2}$$​​

From (2):
y =$$\frac{1}{2}x\left(1+\frac{1}{2}\right) = \frac{3x}{4}$$​​

So,
x : y : z = $$x : \frac{3x}{4} : \frac{x}{2}$$​​

Multiply by 4:
4 : 3 : 2
Required ratio:

​$$\frac{1}{x} : \frac{1}{y} : \frac{1}{z}$$​​
​$$= \frac{1}{4} : \frac{1}{3} : \frac{1}{2}$$​​

Multiply by 12:
3 : 4 : 6