Q.Find the cartesian equation of the plane passing through A(1, 2, 3) and the direction ratios of whose normal are 3, 2, 5.

This is the coordinate-geometry twin of the vector form above. If $A(x_1,y_1,z_1)$ is a known point of the plane and $a, b, c$ are the direction ratios of its normal, then for a general point $P(x,y,z)$ of the plane, the direction ratios of $\overrightarrow{AP}$ are $x-x_1, y-y_1, z-z_1$; because $\overrightarrow{AP}$ is perpendicular to the normal, the sum of the products of corresponding direction ratios vanishes: $a(x-x_1) + b(y-y_1) + c(z-z_1) = 0$. Multiplying out collapses this to the familiar linear form $ax+by+cz+d=0$. …