If you're given a scalar equation
2x-y+3z-24=0
And you're asked to find a vector equation, how can you do this?
There are two ways of expressing the vector equation of a plane:
1. $\displaystyle (\vec{r} - \vec{r_0}) \cdot \vec{n}$ where $\displaystyle \vec{n}$ is a vector normal to the plane and $\displaystyle \vec{r_0}$ is the position vector of a point in the plane.
2. $\displaystyle \vec{r} = \vec{r_0} + s \vec{u} + t \vec{v}$ where $\displaystyle \vec{u}$ and $\displaystyle \vec{v}$ are two non-parallel vectors which lie in the plane and $\displaystyle \vec{r_0}$ is the position vector of a point in the plane. s and t are scalar parameters that can have in real value.
So choose which form of answer you want to give. Then extract the required vectors from the given Cartesian equation and substitute.
A vector normal to the plane ax + by + cz = d is n = ai + bj + ck. You should know this.
You should be able to find a point in the plane and hence get the position vector of this point (=> the position vector $\displaystyle \vec{r_0}$ of a point in the plane).
And, if you want two non-parallel vectors lying in the plane, yuo should be able to find three non-collinear points A, B and C lying in the plane and then construct the vectors u = AB and v = AC, say ....