Statement: Given three planes P1,P2,P3 if their normal vectors n1,n2,n3 form a linearly independent list then given planes intersect at one and only one point.
Proof: Let $U1,U2,U3$ be 2 dimensional subspaces of $R^3$ with $N_j=U_j^\perp;j=1,2,3$ where $N_j=span(n_j)$ $P_j$ is defined as affine subset of $R^3$ parallel to $U_j$,in Cartesian form $P_j=\{{(x_1,x_2,x_3):a_{j1}x_1+a_{j2}x_2+a_{j3}x_3=d_j}\}$ $$U^\perp={\{v \in R^3 : \langle u \, , v \rangle=0 \forall u \in U \}}$$ Let T be an operator defined on $R^3$ as $$T(x1,x2,x3)=(a_{11}x_1+a_{12}x_2+a_{13}x_3,a_{21}x_1+a_{22}x_2+a_{23}x_3,a_{31}x_1+a_{32}x_2+a_{33}x_3)=(d1,d2,d3)$$ Proving above a statement is equivalent to proving T is invertible which is equivalent to proving T is injective which is equivalent to proving U1,U2,U3 has intersection zero subspace.
$U_1\cap U_2=L$ where L is a one dimensional subspace of $R^3$ else $dim(U1+U2)>3$ a contradiction
Consider $L\cap U_3$ ,$dim(L\cap U_3)≤1$ if $$dim(L\cap U_3)=1 \Rightarrow L\cap U_3=L \Rightarrow L\subset U_3 \Rightarrow (U_3)^\perp \subset L^\perp \Rightarrow N_3 \subset L^\perp$$
Let $N=span(n1,n2)$ ,let $n \in N$,consider $l \in L \Rightarrow l \in U_1 ; l \in U_2, \Rightarrow \langle l \, , n \rangle = \langle l \, , c_1n_1+c_2n_2 \rangle= \langle l \, , c_1n_1 \rangle + \langle l \, , c_2n_2 \rangle =0$
Since l is arbitary $\Rightarrow n \in L^\perp \Rightarrow N \subset L^\perp $
Since $Dim(N)=Dim(L^\perp) \Rightarrow N=L^\perp \Rightarrow N_3 \subset N \Rightarrow n3=b_1n_1+b_2n_2$ (contridiction)
$\Rightarrow$ $Dim(L \cap U_3)=0 \Rightarrow L \cap U_3={\{0\}} \Rightarrow U_1 \cap U_2 \cap U_3={0} \Rightarrow T$ is injective $\Rightarrow$ T is invertible hence prove above statement.