Solution: he given PDE is \begin{equation}\label{eq:12July2024-1} \left( \frac{b-c}{a} \right) yzp + \left( \frac{c-a}{b} \right) zxq = \left( \frac{a - b}{c} \right) xy. \end{equation} The corresponding Lagrange's subsidiary equations are \begin{equation}\label{eq:12July2024-2} \frac{a \mathrm{d} x}{(b-c)yz} = \frac{b \mathrm{d} y}{(c-a)zx} = \frac{c \mathrm{d} z}{(a - b) xy} . \end{equation} We recall that if \begin{equation}\label{eq:12July2024-3} \frac{\mathrm{d} x}{P} = \frac{\mathrm{d} y}{Q} = \frac{\mathrm{d} z}{Q} \end{equation} and $P_1, Q_1, R_1$ are functions of $x,y$ and $z$, then \[ \frac{ P_1 \mathrm{d} x + Q_1 \mathrm{d} y + R_1 \mathrm{d} z}{P_1 P + Q_1 Q + R_1 R} \] is equal to each of the fraction of \eqref{eq:12July2024-3}.

Using the equation \eqref{eq:12July2024-2}, and take $P_1 = x, Q_1 = y$ and $R_1 = z$, we have \begin{align*} \frac{a\mathrm{d} x}{(b-c)yz} & = \frac{ax \mathrm{d} x + by \mathrm{d} y + cz \mathrm{d} z}{xyz\left[ (b - c) + (c - a) + (a - b) \right] } . \end{align*} This implies \begin{align*} & ax \mathrm{d} x + by \mathrm{d} y + cz \mathrm{d} z = 0 \\[1ex] \implies & 2\left( ax \mathrm{d} x + by \mathrm{d} y + cz \mathrm{d} z \right) = 0 \\[1ex] \implies & 2ax \mathrm{d} x + 2by \mathrm{d} y + 2cz \mathrm{d} z = 0 \\[1ex] \implies & ax^2 + by^2 + cz^2 = k_1, \end{align*} where $k_1$ is an arbitrary constant.

Again, we can choose $P_1 = ax, Q_1 = by$ and $R_1 = cz$ and each fraction of \eqref{eq:12July2024-2} will be equal to \begin{align*} & \frac{a^2x \mathrm{d} x + b^2y \mathrm{d} y + c^2z \mathrm{d} z}{xyz(a(b-c) + b (c-a) + c(a - b))} = 0\\[2ex] \implies & a^2x \mathrm{d} x + b^2y \mathrm{d} y + c^2z \mathrm{d} z = 0 \\[2ex] \implies & a^2 x^2 + b^2y^2 + c^2 z^2 = k_2, \end{align*} where $k_2$ is an arbitrary constant. Therefore, the solution of the given PDE is \[ \phi (k_1, k_2 ) = 0 \implies \phi \left( ax^2 + by^2 + cz^2 ,a^2 x^2 + b^2y^2 + c^2 z^2\right) = 0 , \] where $\phi$ is an arbitrary function.