1. **State the problem:** We need to find scalars $a$, $b$, and $c$ such that every point $(x,y)$ on the line $ax + by = c$ is equidistant from the points $(2,-1)$ and $(3,2)$. 2. **Recall the definition of equidistant points:** A point $(x,y)$ is equidistant from $(2,-1)$ and $(3,2)$ if the distances to these points are equal: $$\sqrt{(x-2)^2 + (y+1)^2} = \sqrt{(x-3)^2 + (y-2)^2}$$ 3. **Square both sides to eliminate the square roots:** $$ (x-2)^2 + (y+1)^2 = (x-3)^2 + (y-2)^2 $$ 4. **Expand each side:** $$ (x^2 - 4x + 4) + (y^2 + 2y + 1) = (x^2 - 6x + 9) + (y^2 - 4y + 4) $$ 5. **Simplify by canceling $x^2$ and $y^2$ terms:** $$ -4x + 4 + 2y + 1 = -6x + 9 - 4y + 4 $$ 6. **Combine like terms:** $$ -4x + 2y + 5 = -6x - 4y + 13 $$ 7. **Bring all terms to one side:** $$ -4x + 2y + 5 + 6x + 4y - 13 = 0 $$ 8. **Simplify:** $$ ( -4x + 6x ) + ( 2y + 4y ) + (5 - 13) = 0 $$ $$ 2x + 6y - 8 = 0 $$ 9. **Divide entire equation by 2 to simplify:** $$ x + 3y - 4 = 0 $$ 10. **Identify scalars:** Comparing with $ax + by = c$, we have $a = 1$, $b = 3$, and $c = 4$. **Final answer:** $$ a = 1, \quad b = 3, \quad c = 4 $$