How to find the equation of a plane passing through the intersection point of three planes and the origin and another point.

Question

Determine the equation of the plane that passes through the point of intersection of the planes $$P=\begin{cases} 2x + z - 7 = 0\\ x - y = 0\\ x + y - 2z + 2=0 \end{cases}$$ and passes through the origin and the point $Q=(2, 3, 8)$. I tried a lot to find the solution of this problem but in vain.

Answer 1

Here is a quick way of solving this.

The required plane must take the form $$2x+z-7+\lambda(x-y)+\mu(x+y-2z+2)=0$$ For some $\lambda, \mu\in\mathbb{R}$

This is because it is the equation of a plane, and it is satisfied by the point which simultaneously satisfies the equations of all three given planes (i.e. the point of intersection).

Now $(0,0,0)$ lies on the plane, so $\mu=\frac72$.

Also $(2,3,8)$ lies on the plane, so $\lambda=-\frac{53}{2}$

Using these values of $\lambda$ and $\mu$ gives the equation of the plane, which simplifies to $$-7x+10y-2z=0$$

Answer 2

This is the rule used for finding the equation of any plane passing through three coplinear points.