Answer
(a) $x+y+z=11$
(b) $(x-3)=(y-3)=(z-5)$
Work Step by Step
a)
Our aim is to determine the tangent plane equation.
The general form is: $(x_2-x_1)f_x(x_1,y_1,z_1)+(y_2-y_1)f_y(x_1,y_1,z_1)+(z_2-z_1)f_x(x_1,y_1,z_1)=0$ ...(1)
From the given data, we have $f(x,y,z)=(3,3,5)$
Equation (1), becomes:
$(x-3)(4)+(y-3)(4)+(z-5)(4)=0$
or, $4x-12+4y-12+4z-20=0$
Hence, $x+y+z=11$
b)
Our aim is to determine the normal line equation.
The general form is:
$\dfrac{(x_2-x_1)}{f_x(x_1,y_1,z_1)}=\dfrac{(y_2-y_1)}{f_y(x_1,y_1,z_1)}=\dfrac{(z_2-z_1)}{f_x(x_1,y_1,z_1)}$ ...(2)
From the given data, we have $f(x,y,z)=(3,3,5)$
Equation (2), becomes:
$\dfrac{(x-3)}{4}=\dfrac{(y-3)}{4}=\dfrac{(z-5)}{4}$
Hence, $(x-3)=(y-3)=(z-5)$