Q.Find the parametric vector, non-parametric vector and Cartesian form of the equations of the plane passing through the three non-collinear points $(3,6,-2),(-1,-2,6)$, and $(6,4,-2)$.

Equation of a Plane

A plane is fixed by a point on it and a direction perpendicular to it (its normal $\vec n$). Every standard form below is really that one idea written differently.

Point + normal form

If the plane passes through $\vec a$ with normal $\vec n$, then for any point $\vec r$ on it, $\vec r-\vec a$ lies in the plane, so it is perpendicular to $\vec n$:

In Cartesian form with $\vec n=(A,B,C)$: $A(x-x_1)+B(y-y_1)+C(z-z_1)=0$, i.e. $Ax+By+Cz=d$. The coefficients of $x,y,z$ are the normal's direction ratios.

Normal (perpendicular) form

If $\hat n$ is the unit normal and the plane is at distance $p$ from the origin: $\vec r\cdot\hat n=p$, i.e. $lx+my+nz=p$ with $l^2+m^2+n^2=1$.

Intercept form …