(Solved): Notes for the Discussion Point to plane Distance distunce from point P to the plane given by the eq ...

Notes for the Discussion Point to plane Distance distunce from point $P$ to the plane given by the equortion $Ax+By+Cz=d$ is given by $\text{ Distunce }=?\overrightarrow{RP}?\hat{n}?$ where $\hat{n}=\frac{?A,B,C?}{??A,B,C??}$ unit normal vector. and $\overrightarrow{RP}=?x_1?x_0^{?},y_1?y_0,z_1?z_0?$ Why? If you drop a perpendicular line from $P$ to the plane and consider the triangle $PQR$, the distance is the length of the projection of $\overrightarrow{RP}$ onto $\hat{n}$. Find an equation of the plane containing the two pirnkel lines $\vec{r},(t)=?0,1,?2?+t?2,3,?1?$ and ${\vec{r}}_2(t)=?2,?1,0?+t?2,3,?1?$