You are given two planes in parametric form :
\begin{eqnarray*} S_{1} : \begin{pmatrix} x_{1}\\ x_{2}\\ x_{3} \end{pmatrix} = \begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix} + u_1 \begin{pmatrix} 1\\ 2\\ 0 \end{pmatrix} + v_1 \begin{pmatrix} 0\\ 6\\ 1 \end{pmatrix} \end{eqnarray*}
\begin{eqnarray*} S_{2} : \begin{pmatrix} x_{1}\\ x_{2}\\ x_{3} \end{pmatrix} = \begin{pmatrix} 0\\ -1\\ -1 \end{pmatrix} + u_2 \begin{pmatrix} 1\\ 0\\ 3 \end{pmatrix} + v_2 \begin{pmatrix} 0\\ 2\\ 1 \end{pmatrix} \end{eqnarray*}
Substitute expressions for $x_1, x_2$ and $x_3$ from the parametric form of $S_{2}$ into Cartesian equation for $S_1$ and hence find a parametric vector form of the line of intersection $L$.
I found Cartesian equation for $S_1$ which is $2x_1-x_2+6x_3=0$. But when I substitute expressions from $S_2$, they come with $u_2$ and $v_2$ vectors. How can I eliminate them?
Given plane $$S : \vec r = \vec a + u_1 \vec b + u_2 \vec c $$ Alternate way of interpreting this equation as the equation of plane spanned by the vectors $( \vec b - \vec a) = \vec r_1 $ and $(\vec c - \vec a) = \vec r_2$. The normal of the plane can be computed, as $\vec n = \vec r_1 \times \vec r_2 $ and other thing you need is a point which lies on the plane, which here is $\vec a$. Equation of the plane can be written as $ \vec r \cdot \vec n = \vec a \cdot \vec n$. If you will do so, you will obtain the equation of $S_2$ as $10x_1 + 2 x_2 - 3 x_3 = 1$