Converting plane equation from $ax+by+cz=d$ to $r=a+\lambda b+\mu c$

Question

The equation of the plane Π is $$2x + 3y + 4z= 48$$ Obtain a vector equation of Π in the form $r = a + λb + μc$, where a, b and c are of the form pi, qi + rj and si + tk respectively, and where $p, q, r, s, t$ are integers

My Attempt:

$a=(p,0,0)$

$2p+3(0)+4(0)=48 \implies p=24$

---

Then for the other variables I wrote down two equations:

(1) $(2,3,4) \cdot (x,y,z)= (24+\lambda q+ μ s, \lambda r, μ t)$

(2) Cross product of $b$ and $c$ is equal to $(2,3,4)$

This gives $(rt,-qt,rs)=(2,3,4)$

---

It seems impossible to me. Too many variables that my mind can't process. Please help.

Answer 1

Hint:

Let $a$ be a point on the plane, for convenience let's say $a=\begin{pmatrix}24\\0\\0\end{pmatrix}$.

Now, $b$ and $c$ must be vectors that are normal to $\begin{pmatrix}2\\3\\4\end{pmatrix}$, that is for which $\overrightarrow{b}\cdot\begin{pmatrix}2\\3\\4\end{pmatrix}=0$ and $\overrightarrow{c}\cdot\begin{pmatrix}2\\3\\4\end{pmatrix}=0$, where $\cdot$ denotes the dot product.

Answer 2

Hint

In dimension two we have $(a,b)\perp (-b,a).$ This idea can be translated to $3D$ as follows:

$$(a,b,c)\perp (-b,a,0); (a,b,c)\perp (-c,0,a); (a,b,c)\perp (0,-c,b).$$

Can you get from this two vectors perpendicular to the normal vector of the plane $(2,3,4)$ satisfying the required conditions?