Q.Find the vector and cartesian equations of the plane through the points $(1, 2, 3)$ and $(2, 3, 1)$ and perpendicular to the plane $3x - 2y + 4z - 5 = 0$.

Find the plane's normal as the cross product of (the line joining the two given points) and (the normal of the given perpendicular plane), then fit it through one of the points.

1. Let the required plane have equation $a(x-1)+b(y-2)+c(z-3)=0$ (it passes through $(1,2,3)$, with normal $\vec n=(a,b,c)$).
2. Since the plane also passes through $(2,3,1)$: $a(2-1)+b(3-2)+c(1-3)=0\Rightarrow a+b-2c=0$ — so $\vec n$ is perpendicular to the direction $\vec d_1=(1,1,-2)$ joining the two points.
3. Since the required plane is perpendicular to $3x-2y+4z-5=0$ (normal $\vec d_2=(3,-2,4)$), the two planes' normals are themselves perpendicular: $\vec n\cdot\vec d_2=0\Rightarrow3a-2b+4c=0$.
4. Since $\vec n$ is perpendicular to both $\vec d_1=(1,1,-2)$ and $\vec d_2=(3,-2,4)$, take $\vec n=\vec d_1\times\vec d_2$.
5. $\vec d_1\times\vec d_2=\begin{vmatrix}\vec i&\vec j&\vec k\\1&1&-2\\3&-2&4\end{vmatrix}=\vec i[(1)(4)-(-2)(-2)]-\vec j[(1)(4)-(-2)(3)]+\vec k[(1)(-2)-(1)(3)]$. …