Calculus intersection and equation of planes with vectors I am attempting to do this problem for my calc homework and I am stuck on trying to set up the equations to find the intersection point. I understand that the equations that are given to me are in the form of a vector equation, $v = r + tv$ where $r$ and $v$ are vectors. I'm not exactly sure how to get the points and the equation from these vector equations tho.
Consider the lines below.
$r = \langle 2, 3, 0\rangle + t \langle3, -3, 3\rangle$
$r = \langle5, 0, 3\rangle + s\langle-3, 3, 0\rangle$
(a) Find the point at which the given lines intersect.
$(x= ,y= ,z= )$
(b) Find an equation of the plane that contains these lines.
 A: Set the two equations equal to each other. Then for each of the three coordinates, you will get an equation in $s$ and $t$, so you will have three equations in two variables. If the two lines are co-planer then there is a unique solution for $s$ and $t$ which will give you the coordinates of the point of intersection. 
For example, the first coordinates give you the equation
$$ 2+3t=5-3s $$
Find the equations for the other two coordinates and finish the problem.
ADDENDUM:
Now that you correctly found the point of intersection $(5,0,3)$ you have the necessary information to find the equation of the plane which contains the two intersecting lines.
To find the equation of a plane containing two intersecting lines you need three pieces of information: direction vectors for each of the two lines and the point of intersection of the two lines.
The direction vectors are the vector coefficients of your two vector line equations:


*

*$\langle 3,-3,3\rangle$

*$\langle 3,-3,0\rangle$


These two may be simplified by multiplying by $\dfrac{1}{3}$ since multiplication by a nonzero constant does not change the direction of a vector. So use the following for the two direction vectors.


*

*$\langle 1,-1,1\rangle$

*$\langle 1,-1,0\rangle$


The cross-product of these two vectors gives a normal vector $N=\langle a,b,c\rangle$ for the plane containing the two lines.
$N=\begin{vmatrix}
\mathbf{i}&\phantom{-}\mathbf{j}&\mathbf{k}\\1&-1&1\\1&-1&0\end{vmatrix}=\mathbf{i}+\mathbf{j}=\langle 1,1,0\rangle$
The equation of a plane has the form
$$ ax+by+cz=d $$
where the normal vector is $\langle a,b,c\rangle$ and $d$ is computed using the values of $a,b,c$ and the coordinates of a point in the plane ( in this case, $(5,0,3)$).
So the equation for this plane is
$$ 1\cdot x+1\cdot y+0\cdot z=d$$
where
$$1(5)+1(0)+0(3)=d=5$$
So the equation of the line containing the two lines is
$$ x+y=5 $$
