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Line equation through point, parallel to plane and intersecting line

Write the equations of the line that passes through point $P(1,-2,-1)$, is parallel with the plane $x+3y-z=0$ and intersects line $(x,y,z)=(1+t,2-t,3+2t)$ I cant come about to solve it.

Any help appreciated!!

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    I know that the dot product of the line and and the plane must be zero ! and the two lines must have a common point when they intersect. so i plugged the line equation in the plane equation to solve for (t) and i did get the points(x,y,z) but then when i did the dot product i did not get zoro !! @TheCount – 2017-02-23
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    One way to solve this could be to determine the plane $\alpha$ containing $P$ and parallel to the given plane. Then you determine the intersection of $\alpha$ and the given line, which will be a point (unless the line lies in $\alpha$). If it is a piont: take the line through the intersection point and $P$. If is not a point: take any line through $P$, in $\alpha$ and not parallel to the given line. – 2017-02-23

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If the helping guide in my comment was not sufficient, I have turned it into a solution, which should give you some insight in how to cope with these questions.

The plane $\alpha$ parallel to the plane determined by the points satisfying $x + 3y - z = 0$ and containing the point $P$ is given by
$$\alpha \leftrightarrow (x - 1) + 3 (y +2) - (z +1) = 0.$$

Let us now rewrite the equation of the given line. Since a line is $1$ dimensional, the points on the line will be the solution of a system containing two linearly independent equations. We have that \begin{equation} \begin{cases} x = 1+t\\ y = 2 -t\\ z = 3 + 2t \end{cases} \end{equation} and adding the first two equations gives us $x + y = 3$, whereas adding two times the second equation to the third equation gives $2y + z = 7$. Therefore the line $l$ is given by \begin{equation} l \leftrightarrow \begin{cases} x + y = 3\\ 2y + z = 7 \end{cases} \end{equation} so if we want to find the intersection of $\alpha$ with $l$, we need to solve the following system of equations: \begin{equation} \begin{cases} x + 3y - z = -4\\ x + y = 3\\ 2y + z = 7 \end{cases} \end{equation} (do you see where the first equation comes from?). This can be done by writing this as an augmented matrix and row reducing it (I'll leave this to you). You should find that the solution is $Q = (3,0,7)$. Hence the line we are looking for is the line passing through $P$ and $Q$. The direction of this line is the vector $\vec{PQ}$, which is given by $Q - P = (2,2,8)$. So your line has parameter equation $P + t(2,2,8) = (1 + 2t, -2 + 2t, -1 + 8t)$, for $t \in \mathbb{R}$, which you can rewrite as you wish.

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    Thank u vary much !! now i understand !! :D @ Student – 2017-02-23
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    @haidar: if you understand, please consider to accept my answer by highlighting the green checkmark. This shows thatyour question is solved. – 2017-02-23