Answer
$L(1,0,0) =y-3z$
and $L(1,1,0) =x+y-z-1$
Work Step by Step
$f_x=y-3z \\ \implies f_x(1,0,0)=0$
Next, $f_y =x+2z \\ \implies f_{y}(1,0,0) = 1$ and $f_y =2y-3x \\ \implies f_{z}(1,0,0) = -3$
Now, a) $L(1,0,0) =0(x-1)+(1)(y-0) -3(z-0)\\ =y-3z$
and $f_x=y-3z \\ \implies f_x(1,1,0)=1$
Next, b) $f_y =x+2z \\ \implies f_{y}(1,1,0) = 1$ and $f_z =2y-3x \\ \implies f_{z}(1,1,0) = -1$
So, $L(1,1,0) =1 (x-1)+(1)(y-1) -1 (z-0)\\ =x+y-z-1$
