I tried to make the question as brief as possible in the title without omitting critical aspects.
There are three elements to it.
1) A position vector [xyz] is given first and then
2) Its perpendicular plane must go through position [xyz], and
3) Lastly find the intercepts the plane makes with the XYZ axes
The vectors coming in are arbitrary, but we may restrict them to positive quadrant of the coordinate axes in the answer.
With that quadrant restriction the plane completes a solid triangle with the intercepts forming the base corners, and origin the apex corner. But in my case the intercepts may be any length so long as constraints [1] [2] are met.
I have not found a standard procedure out there (yet)
Thanks In Advance
I assume You are referring to a plane which pass through a point and is perpendicular to that point’s position vector.
If the point we were talking about is $\vec{r_{0}}$, then the plane is $\vec{r_{0}}\cdot\left(\vec{r_{0}}-\vec{r}\right)=0$.
To find intercept with $x$ - axis, simply make $\hat{j}$ and $\hat{k}$ component of $\vec{r}$ as zero and try to find its $\hat{i}$ component. For $y$ and $z$ axis, i am sure you can manage now. I will give an example below.
Example
Find a plane passing through $(1,2,3)$ and is perpendicular to its position vector, then find its intercept with the axes.
$$ \begin{aligned} \left(1\hat{i}+2\hat{j}+3\hat{k}\right)\cdot\left((1-x)\hat{i}+(2-y)\hat{j}+(3-z)\hat{k}\right)&=0\\ 1-x+4-2y+9-3z&=0\\ 14&=x+2y+3z \end{aligned} $$
To find intercept with $x$ axis, set $y=z=0$
$$ 14=x $$
To find intercept with $y$ and $z$ axes, similar
$$ \begin{aligned} 14&=2y\\ 7&=y\\ \\ 14&=3z\\ \frac{14}{3}&=z \end{aligned} $$