Answer
(a) $x+2y+6z=12$
(b) $(x-2)=\dfrac{(y-2)}{2}=\dfrac{(z-1)}{6}$
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$(2,2,1)$
$(x-2)(4)+(y-2)(8)+(z-1)(24)=0$
$4x-8+8y-16+24z-24=0$
This implies, $4x+8y+24z=48$ or, $x+2y+6z=12$
(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$(2,2,1)$
$\dfrac{(x-2)}{4}=\dfrac{(y-2)}{8}=\dfrac{(z-1)}{24}$
or, $(x-2)=\dfrac{(y-2)}{2}=\dfrac{(z-1)}{6}$
