Answer
(a) $x+y+z=11$
(b) $(x-3)=(y-3)=(z-5)$
Work Step by Step
(a) Formula to calculate tangent plane equation is:
$(x_2-x_1)f_x(x_1,y_1,z_1)+(y_2-y_1)f_y(x_1,y_1,z_1)+(z_2-z_1)f_x(x_1,y_1,z_1)=0$
At point$(3,3,5)$
$(x-3)(4)+(y-3)(4)+(z-5)(4)=0$
$4x-12+4y-12+4z-20=0$
This implies, $x+y+z=11$
(b) Formula to normal line equation is:
$\dfrac{(x_2-x_1)}{f_x(x_1,y_1,z_1)}=\dfrac{(y_2-y_1)}{f_y(x_1,y_1,z_1)}=\dfrac{(z_2-z_1)}{f_x(x_1,y_1,z_1)}$
At point$(3,3,5)$
$\dfrac{(x-3)}{4}=\dfrac{(y-3)}{4}=\dfrac{(z-5)}{4}$
or, $(x-3)=(y-3)=(z-5)$
