#### Answer

(a) $F_xG_x+F_yG_y+F_zG_z=0$ at $P$
(b) The two surfaces are orthogonal and their dot product is equal to zero.

#### Work Step by Step

a)
Condition to be orthogonal: Two normal lines are known to be orthogonal at point P when $\nabla F \cdot \nabla G=0$ at point $P$ or, we can also say that their dot product must be zero.
$\lt \dfrac{\partial F}{\partial x},\dfrac{\partial F}{\partial y},\dfrac{\partial F}{\partial z} \gt \cdot \lt \dfrac{\partial G}{\partial x},\dfrac{\partial G}{\partial y},\dfrac{\partial G}{\partial z} \gt =0$
Thus, $\lt \dfrac{\partial F}{\partial x}\dfrac{\partial G}{\partial x},\dfrac{\partial F}{\partial y}\dfrac{\partial G}{\partial y},\dfrac{\partial F}{\partial z} \dfrac{\partial G}{\partial z} \gt =0$
Hence, $F_xG_x+F_yG_y+F_zG_z=0$ at $P$
b)
From the given data in the question, we have
$\nabla P(x,y,z) \cdot \nabla Q(x,y,z)=\lt 2x,2y,-2z \gt \times \lt 2x,2y,2z \gt=4(x^2+y^2-z^2)$
or, $\nabla P(x,y,z) \cdot \nabla Q(x,y,z)=0$
Hence, we can see that the two surfaces are orthogonal and their dot product is zero.