If ${\mathrm{P}}_1:\overrightarrow{\mathrm{r}}\cdot {\overrightarrow{\mathrm{n}}}_1-{\mathrm{d}}_1=0,{\mathrm{P}}_2:\overrightarrow{\mathrm{r}}\cdot \overrightarrow{{\mathrm{n}}_2}-{\mathrm{d}}_2=0$ and  ${\mathrm{P}}_3:\overrightarrow{\mathrm{r}}\cdot \overrightarrow{{\mathrm{n}}_3}-{\mathrm{d}}_3=0$ are three planes $\overrightarrow{{\mathrm{n}}_1},\overrightarrow{{\mathrm{n}}_2}\text{ and }\overrightarrow{{\mathrm{n}}_3}$ are  three non-coplanar vectors, then three lines P₁ =0, P₂=0; P₂= 0, P₃ = 0 and P₃ : 0, P₁ : 0 are

P₁ =0, P₂=0; P₂= 0, P₃ = 0 and P₃ : 0, P₁ : 0 are  lines of intersection of the three planes P ₁ P₂ and P₃

As $\overrightarrow{{\mathrm{n}}_1},\overrightarrow{{\mathrm{n}}_2}\text{ and }\overrightarrow{{\mathrm{n}}_3}$  are non-coplanar, planes P₁, P₂ and P₃ will intersect at unique point. 

So the given lines will pass through a fixed point.