Given $$a, b, c$$ are real numbers such that $$9a+b+8c=12$$ and $$8a-12b-9c=1$$. Then $$a^2-b^2+c^2=$$

Combining the two equations eliminates $$b$$ to give $$4a+3c=5$$, and the direction vector of the resulting solution line is $$(-3,-5,4)$$, whose components satisfy $$(-3)^2-(-5)^2+4^2=9-25+16=0$$. This means $$a^2-b^2+c^2$$ is constant along all solutions, and evaluating at one solution (such as $$a=1.25, b=0.75, c=0$$) gives the value 1.