# Science:Math Exam Resources/Courses/MATH152/April 2015/Question B 3 (a)/Solution 2

Let ${\displaystyle \mathbf {x} =(x_{1},x_{2},x_{3})}$ denote the point of intersection. Since ${\displaystyle \mathbf {x} \in P_{1}}$, we must have
${\displaystyle {\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\end{bmatrix}}={\begin{bmatrix}1\\1\\1\end{bmatrix}}+s_{1}{\begin{bmatrix}0\\1\\0\end{bmatrix}}+s_{2}{\begin{bmatrix}1\\2\\-1\end{bmatrix}}={\begin{bmatrix}1+s_{2}\\1+s_{1}+2s_{2}\\1-s_{2}\end{bmatrix}}.}$
In particular, ${\displaystyle x_{1}+x_{3}=2}$. On the other hand, since ${\displaystyle \mathbf {x} \in l_{1}}$, we must also have
${\displaystyle {\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\end{bmatrix}}=t{\begin{bmatrix}0\\1\\2\end{bmatrix}}={\begin{bmatrix}0\\t\\2t\end{bmatrix}}.}$
It follows that ${\displaystyle 0+2t=2}$, so ${\displaystyle t=1}$. We conclude that
${\displaystyle \mathbf {x} ={\color {blue}{\begin{bmatrix}0\\1\\2\end{bmatrix}}}.}$