Answer
(a) $3x+2y+3z=10$
(b) $\dfrac{(x-1)}{3}=\dfrac{(y-2)}{2}=\dfrac{(z-1)}{3}$
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$(1,2,1)$
$(x-1)(3)+(y-2)(2)+(z-1)(3)=0$
$3x-3+2y-2+3z-3=0$
$5(3x-3+2y-2+3z-3)=0$
This implies, $3x+2y+3z=10$
(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$(1,2,1)$
$\dfrac{(x-1)}{3}=\dfrac{(y-2)}{2}=\dfrac{(z-1)}{3}$
