#### Answer

$z=x-y+1$

#### Work Step by Step

Given: $z=e^{x-y}$, $(2,2,1)$
Consider $f(x,y)= e^{x-y}$
$f_{x}(x,y)=e^{x-y}$
$f_{y}(x,y)=-e^{x-y}$
At $(2,2,1)$
$f_{x}(2,2)=e^{2-2}=1$
$f_{y}(2,2)=-e^{2-2}=-1$
The equation of the tangent plane to the given surface at the specified point $(2,2,1)$ is given by
$z-z_{0}=f_{x}(x_{0},y_{0})(x-x_{0})+f_{y}(x_{0},y_{0})(y-y_{0})$
On substituting the values, we get
$z-1=1(x-2)-1(y-2)$
$z=x-2-y+2+1$
Hence, $z=x-y+1$