Systems of equations

With $A = 17$ , $B = 34$ , and $C = 87$ , we have

$Ax+sy+tz = 0 \\ rx+By+tz = 0 \\ rx+sy+Cz = 0$

Since it is specified that $r \neq A$ and $x \neq 0$ , we know that $Ax \neq rx$ . Now if $s$ were $B$ , both $Ax$ and $rx$ would need to equal $(-By + tz)$ , which is impossible. So $s \neq B$ . Similarly, we get $t \neq C$ , so the divisions in the sought quantity make sense.

Because $x \neq 0$ , the determinant

$\left|\begin{matrix}A & s & t \\ r & B & t \\ r & s & C \end{matrix}\right|$

needs to be $0$ . We can compute this determinant directly using Sarrus's rule:

$\tag{1} 0 = ABC + 2rst - Ast - rBt - rsC$

The quantity we want to find is

$\displaystyle \frac{A}{A - r}+\frac{B}{B - s}+\frac{C}{C - t}$

which, solving for a common denominator, is

$\displaystyle \frac{A(B - s)(C - t) + (A - r)B(C - t) + (A - r)(B - s)C}{(A - r)(B - s)(C - t)}$

Multiplying out and collecting like terms in the numerator and denominator, this equals

$\displaystyle \frac{3ABC - 2(ABt + AsC + rBC) + (Ast + rBt + rsC)}{ABC - (ABt + AsC + rBC) + (Ast + rBt + rsC) - rst}$

Change one of the $ABC$ s in the numerator to $(Ast+rBt+rsC)-2rst$ using $(1)$ , and we get

$\displaystyle \small \frac{2ABC - 2(ABt + AsC + rBC) + 2(Ast + rBt + rsC) - 2rst}{ABC - (ABt + AsC + rBC) + (Ast + rBt + rsC) - rst} = \boxed{2}$