Let $X, Y, Z$ be linear varieties of dimension $r, s, t$ respectively in $\mathbb{P}^n$.
If $r+s\ge n$, then $X\cap Y\neq \varnothing$. Furthermore, if $X\cap Y\neq \varnothing$, then $X\cap Y$ is a linear variety of dimension $\ge r+s-n$. (Hartshorne's Algebraic Geometry, Chap1, Ex2.11)
My questions are:
Suppose $r+s+t\ge n$. Is it possible that $X\cap Y=\varnothing$, $Y\cap Z=\varnothing$ and $Z\cap X=\varnothing$? I think the intersection of any two of $X, Y, Z$ is empty only if $(r+1)+(s+1)+(t+1)<(n+1)$, but I can't confirm it. Is it right?
Suppose $r=2, s=1, t=1$ and $n=3$. (i.e. $X$ a plane, $Y,Z$ lines in $\mathbb{P}_{<x,y,z,w>}^3$). Let $P,Q,R$ be three distinct points. If $X\cap Y=P$, $Y\cap Z=Q$ and $Z\cap X=R$, then can I assume $X:w=0, Y:x=y=0, Z:y=z=0$ s.t. $P=(0,0,1,0), Q=(0,0,0,1), R=(1,0,0,0)$ after coordinate change?
For part 1, you'll need $r+s$, $s+t$, and $r+t$ to all be less than $n$ for each pairwise intersection to be empty. Adding these equations, you see that $2(r+s+t)<3n$, and combining with your restriction $r+s+t\geq n$, we have $2n\leq 2(r+s+t)<3n$. Certainly solutions to this inequality exist- let $n=5, r=s=t=2$. So we can pick $X=(x_0,x_1,x_2)$, $Y=(x_3,x_4,x_5)$, $Z=(x_3-x_0,x_4-x_1,x_5-x_2)\subset \mathbb{P}^5=\operatorname{Proj}k[x_0,x_1,x_2,x_3,x_4,x_5]$ and these do the trick.
For part 2, go to the affine cone to make this into a bit of linear algebra. Let $\hat{X}$ be the affine cone over $X$. Then this asks if a 3-plane and two 2-planes which all have pairwise intersections in lines can be coordinate changed to the situation you gave. I think I might have a counterexample, but it is late where I am and I cannot think of it right now. I'll come back to this.