<s>[INST] vectorial equation i need help to Calculate the vectorial equation of the line $$ g = E_{1}\cap E_{2} $$ you get by intersecting the planes$$ E_{1}:\overrightarrow{x}=\left(\begin{array}{c}1\\ 0\\1\end{array}\right)+r.\left(\begin{array}{c}1\\ 1\\0\end{array}\right)+s.\left(\begin{array}{c}0\\ 1\\1\end{array}\right);r,s\in R $$ $$ E_{2}:\overrightarrow{x}=\left(\begin{array}{c}1\\ 1\\1\end{array}\right)+v.\left(\begin{array}{c}1\\ 1\\1\end{array}\right)+w.\left(\begin{array}{c}1\\ 0\\0\end{array}\right);v,w\in R $$
And to determine the vectorial equation of the line ( ℎ ), so that ( ℎ ) is perpendicular to ( E1 ) and that the point A(1|0|1) is located on ℎ $$ ((h \bot E_{1}) ; A(1|0|1)\in h) $$
 [/INST] The normal vectors to planes $E_1$ and $E_2$ are:
$$\vec{N_1}=\begin{pmatrix}1\\1\\0\end{pmatrix} \times \begin{pmatrix}1\\1\\0\end{pmatrix}=\begin{pmatrix}1\\-1\\1\end{pmatrix}  \ \text{and} \ \vec{N_2}=\begin{pmatrix}1\\1\\1\end{pmatrix} \times \begin{pmatrix}1\\0\\0\end{pmatrix}=\begin{pmatrix}0\\1\\-1\end{pmatrix} \ \text{resp.}$$
(They are attached to the vectorial planes associated with these affine planes).
Thus, a directing vector of the intersection line is:
$$\vec{N}:=\vec{N_1}\times\vec{N_2}=\begin{pmatrix}0\\1\\-1\end{pmatrix}$$
Now let us find a common point of the affine planes. This can be done by inspection: $A=\begin{pmatrix}2\\1\\1\end{pmatrix}$ is such a point (take $r=1,s=0,v=0,w=1$).
Thus a parametric representation of the looked for intersection line :
$$M=A+t\vec{N}=\begin{pmatrix}2\\1\\1\end{pmatrix}+t\begin{pmatrix}0\\1\\-1\end{pmatrix}=\begin{pmatrix}2\\1+t\\1-t\end{pmatrix} \ \ t \in \mathbb{R}$$
Edit: it is interesting to check the answer by writing that intersection line $g$ lies in $E_1$ for example:
$$\begin{cases}2&=&1+r\\1+t&=&r+s\\1-t&=&1+s\end{cases}$$
which are compatible with $r=1$ and $s=-t$. In particular $r=1$ can be considered as the equation of $g$ in plane $E_1$. 
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