​​$$\text{The general solution of the surfaces which are} \\ \text{perpendicular to the family of surfaces} \\z^2 = kxy, \; k \in \mathbb{R} \ \text{is}$$​​

Correct option is C

​$$\text{Given surface is } z^2 = K \cdot x \cdot y \implies \frac{xy}{z^2} = K \\\text{so, } f(x, y, z) = \frac{xy}{z^2} \\\implies f_x = \frac{y}{z^2}, \quad f_y = \frac{x}{z^2}, \quad f_z = -\frac{2xy}{z^3} \\\text{Equation of surfaces perpendicular to the given surface has p.d.e as } \\f_x \cdot p + f_y \cdot q + f_z \cdot (-1) = 0 \\\implies \frac{y}{z^2} \cdot p + \frac{x}{z^2} \cdot q + \frac{2xy}{z^3} = 0 \\\implies y \cdot p + x \cdot q = -\frac{2xy}{z} \\\text{So, it's Lagrange's auxiliary equation is } \frac{dx}{y} = \frac{dy}{x} = \frac{z \, dz}{-2xy} \\\text{From the first two equations, we get } x^2 - y^2 = C_1$$

$$\text{From the first and third equation, we get } z \, dz + 2x \, dx \implies z^2 + 2x^2 = C_2. \\\text{So the general solution is :}\\ \phi(x^2 - y^2,z^2+2 x^2) = 0.$$​​​