My 800 followers problem!

Let $x-y=p \quad , \quad y-z=q \quad ,\quad z-x=r$ , then the system gets transformed to :

$\begin{array}{c}(p & + & q & + & r &=0 & \dots (1)\\ 799p& +& 800q& + &801r & =0 & \dots (2) \\ 799^2p & + & 800^2q & +& 801^2r & =800 &\dots (3) \end{array}$

$[800\times(eqn.(1))] - (eqn.(2))\Rightarrow p-r=0 \Rightarrow p=r \dots (4)$

$\text{Substituting (4) in (1) } \Rightarrow q=-2p \dots (5)$

$\text{Substituting (4),(5) in (3)} \Rightarrow 799^2p+800^2(-2p)+801^2(p)=800 \\ \Rightarrow p(799^2-(2)800^2 + 801^2)=800 \\ \Rightarrow p[(799+800)(799-800) + (801+800)(801-800)]=800 \\ \Rightarrow p(-1599+1601)=800 \\ \Rightarrow 2p=800\\ \Rightarrow p = 400 \quad,\quad q=-800$

$\Rightarrow x-2y+z = x-y-(y-z) = p-q=400-(-800)=400+800 \\ \Rightarrow x-2y+z=\huge\boxed{\color{#D61F06}{1200}}$

Bonus: Discover that $\quad y,x,z$ , in this order form an Arithmetic Progression.

There is an alternative way to solve this,Eqs can be written as, $-2x+y+z=0$ $-3200x+1599y+1601z-800=0$ Let $x-2 y+z=M$ or $x-2 y+z-M =0$ As only 2 eqs are given for 3 variable but we have to find value of M,so above 3 eqs must be linearly dependent so let, $-3200x+1599y+1601z-800=p(-2x+y+z)+q( x-2y+z-M)$ Compare coefficient & constant, $qM=800$ $-2p+q=-3200$ $p-2q=1599$ $p+q=1601$ Above 3 eqs have unique soln (3p , 3q)=(4801 , 2) $M=800/q=2400/3q=2400/2=1200$