The Decision of three methods of the solutions $dx/P=dy/Q=dz/R$ Question: 
(A) $$\frac{adx}{(b-c)yz}= \frac{bdy}{(c-a)xz}=\frac{cdz}{(a-b)xy}$$
(B) $$\frac{dx}{xz-y}=\frac{dy}{yz-x}=\frac{dz}{1-z^2}$$

These are simultaneous diff eq. of the first order and the first degree in three variables
And I know that three methods of solution of $dx/P=dy/Q=dz/R$ 
First method: 

Second method: 

Third method: 

I know them. But I cannot decide which method I need to use for above two questionss. I dont want to solve these two questions. Only I want to learn which methods I use respectively? Please give me suggestion or hint? Thank a lot. 

A.
$$\frac{a dx}{(b -c ) yz} = \frac{b dy}{(c -a ) xz} = \frac{c dx}{(a -b ) xy}$$
Now multiply the numerator and denominator first fraction by $x$ for second fraction by $y$ and third fraction by $z$ and add all of them and get it 
$$\frac{ax dx}{(b-c)xyz} = \frac{by dy}{(c-a)xz} = \frac{cz dz}{(a-b)xy} = \frac{ax dx + by dy + cz dz}{(b-c+c-a + a - b) xyz} $$
So you shall get $ax dx + by dy + cz dz = 0$ from the last fraction. One solution $a\frac{x^2}{2} + b\frac{y^2}{2} + c\frac{z^2}{2} =\text{constant}$ will comes from here. 
For other solution take any two of the fractions such as 
$$\frac{adx}{(b-c)yz} = \frac{bdy}{(c - a)xz}$$
and see one term will be cancelled from the denominator.
So you shall get 
$a(c-a)x dx = b(b-c)y dy$
i.e. $a(c-a)\frac{x^2}{2} - b(b-c)\frac{y^2}{2}$ = constant.
I hope one shall get it easily.
B.
$$\frac{dx}{xz - y} = \frac{dy}{yz - x} = \frac{dz}{1 - z^2}$$
Do it as follows.
$$\frac{dz}{1 - z^2} = \frac{dx + dy + 0 dz}{(x + y)z - (x+y) + 0. (1 - z^2)}$$
Simplify it and get one solution from the relation $\frac{dz}{1 - z^2} = \frac{d(x+y)}{x+y}$.
For he other solution take $dx - dy + 0dz$ instead of $dx + dy + 0 dz$. 
I hope everybody will be satisfied.
