#### Answer

$z=3x+7y-5$

#### Work Step by Step

$z=x^{2}+xy+3y^{2}$, $(1,1,5)$
Consider $f(x,y)= x^{2}+xy+3y^{2}$
$f_{x}(x,y)=2x+y$
$f_{y}(x,y)=x+6y$
At $(1,1,5)$
$f_{x}(1,1)=2.1+1=3$
$f_{y}(1,1)=1+6.1=7y$
The equation of the tangent plane to the given surface at the specified point $(1,1,5)$ 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
$z-5=3(x-1)+7(y-1)$
$z=3x-3+7y-7+5$
Hence,$z=3x+7y-5$