I can't figure out a geometric interpretation of what this question wants. I thought maybe it was akin to a line of best fit but I am not sure. An d $f(x_1)$ the value of the function at $x_1$, but isn't that the same as $y_1$, so wouldn't $f(x_1)-y_1=0?$
Let us consider that the $x$-axis is horizontal and that the $y$-axis is vertical. Here, we know a set of 3 vertices $(x_i,y_i)_{i=1\dots3}$ of the plane $\mathbb{R}^2$ (red points on the figure). We want to find the two parameters $a$ and $b$ of the function $f:x\mapsto ax+b$ (black line) that minimize the $\infty$-norm of the difference between the red dots $(x_i,y_i)_{i=1\dots3}$ and the blue squares $(x_i,f(x_i))_{i=1\dots3}$. So yes, it consists in finding the best-fit line in $\infty$-norm.
Mathematically, the problem to solve is finding $(a^*,b^*)$ such that $$ (a^*,b^*) = \underset{(a,b)\in\mathbb{R}^2}{\arg\min}\, \|a \boldsymbol{x} + b - \boldsymbol{y}\|_{\infty} \, , $$ where $\boldsymbol{x} = (x_1,x_2,x_3)$ and $\boldsymbol{y} = (y_1,y_2,y_3)$, i.e. $$ (a^*,b^*) = \underset{(a,b)\in\mathbb{R}^2}{\arg\min}\, \max_{i\in\lbrace 1\dots3\rbrace} |a x_i + b - y_i| \, . $$
