Answer
(a) $8x-y-6z=7$
(b) $\dfrac{(x-4)}{-8}=\dfrac{(y-7)}{1}=\dfrac{(z-3)}{6}$
or, $\dfrac{(x-4)}{8}=-\dfrac{(y-7)}{1}=-\dfrac{(z-3)}{6}$
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$(4,7,3)$
$(x-4)(-8)+(y-7)(1)+(z-3)(6)=0$
$-8x+32+y-7+6z-18=0$
This implies, $8x-y-6z=7$
(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$(4,7,3)$
$\dfrac{(x-4)}{-8}=\dfrac{(y-7)}{1}=\dfrac{(z-3)}{6}$
or, $\dfrac{(x-4)}{8}=-\dfrac{(y-7)}{1}=-\dfrac{(z-3)}{6}$
