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