find a plane that passes through one line and is parallel to another This is how the question was given to me:

Find the equation of a plane that passes through the line $$p: \frac {x-3}{2} = \frac {y+4}{1}= \frac{z-2}{-3}$$
  and is parallel to another line $$ q: \frac {x+5}{4} = \frac {y-2}{7}= \frac{z-1}{2} $$

So I know if a plane is parallel to a line that means the dot product of the normal vector of that plane and the vector given by the line $q=(4,7,2)$ should be zero. So that gives me one equation: $4x+7y+2z=0$.
And if the line p passes through the plane then by substituting the equations given for $x$, $y$, $z$ (from the parametric equations: $p:x=2t+3; y=t-4; z=-3+2$) I can get another equation: $A(2t+3)+B(t-4)+C(-3t+2)$
If I combine the two I get that $t=\frac {12}{9}$ which doesn't seem right. Also I have no idea what to do further than that. Where did I go wrong? 
 A: Probably, the best approach is that you find first a point $P$ on $p$ ( simply put e.g. $z=2$ and get $y$ and $x$ so $P=(3,-4,2)$).
Then construct the equation of a plane passing through $P$ with normal vector equal to the cross-product of the direction vectors of $p$ and $q$ :$(23,-16,10)$. 
That is  
$$
23\left( {x - 5} \right) - 16\left( {y + 3} \right) + 10\left( {z + 1} \right) = 0\quad  \Rightarrow \quad 23x - 16y + 10z - 153 = 0
$$
and if you substitute in it the parametric equations for $p$
$$
x = 2t + 3,\quad {\rm  }y = t - 4,\quad {\rm  }z =  - 3t + 2
$$
you get $0=0$, and of course you have
$$
\eqalign{
  & 23 \cdot 2 - 16 \cdot 1 + 10 \cdot \left( { - 3} \right) = 0  \cr 
  & 23 \cdot 4 - 16 \cdot 7 + 10 \cdot 2 = 0 \cr} 
$$
Going your way, a general plane parallel to $q$ is $q_x(x-x_0)+ \cdots =0 $, i.e. $q_x x+ \cdots=d$.
If in it you put the parametric equations (in $t$) of $p$, since $p$ lies on the plane, the plane equation shall be satisfied for whichever value of $t$, which means that you shall find the $d$ which leads you to $0t=0$.
A: use that the plane has the equation
$$ax+by+cz+d=0$$ and since the line should be parallel to the plane the dot-product of $$\vec{n}=[a;b;c]$$ and the direction vector of the second line $$\vec{a}=[4;7;2]$$ must be zero and we get
$$4a+7b+2c=0$$
since the plane containes the first line we get
$$a(2t+3)+b(t+4)+c(-3t+2)+d=0$$
simplifying this equation
$$t(2a+b-3c)+3a+4b+2c+d=0$$
then we get $$2a+b-3c=0$$ and $$3a+4b+2c+d=0$$
can you finish?
A: Note that


*

*line $p$ has equation $(3,-4,2)+t(2,1,-3)$

*line $q$ has equation $(-5,2,1)+t(4,7,2)$


Thus a plane containing $p$ has equation $ax+by+cz+d=0$ such that $2a+b-3c=0$ that is


*

*$ax+(3c-2a)y+cz+d=0$


From the condition that it contains the point $(3,-4,2)$we obtain


*

*$3a-12c+8a+2c+d=0\implies d=10c-11a$

*$ax+(3c-2a)y+cz+10c-11a=0\implies x+(3k-2)y+kz+10k-11=0$
to be parallel to plane $q$ normal vector must be orthogonal to direction vector of $q$ that is


*

*$4+21k-14+2k=0\implies k=\frac{10}{23}$


therefore
$$x-\frac{16}{23}y+\frac{10}{23}z+\frac{100}{23}-11=0\iff23x-16y+10z-153=0$$
In short the plane equation $ax+by+cz+d=0$ can be obtained assuming $(a,b,c)=v_p\times v_q$ and finding $d$ by the condition that $(3,-4,2)$ belongs to it.
A: This is pretty straightforward if you take a projective-geometric approach and use homogeneous coordinates. A parameterization of the first line is $(3,-4,2)+\lambda(2,1,-3)$. The homogeneous coordinates of two points on the line can be read directly from this parameterization (and also from the original equations) $[3:-4:2:1]$ and $[2:1:-3:0]$. The latter is a point at infinity that corresponds to the line’s direction vector. The plane must be parallel to $q$, which means that their intersection is the point at infinity that corresponds to $q$’s direction vector, so we have a third point on the plane, $[4:7:2:0]$.  
For any point $\mathbf x$ on the plane $\mathbf\pi$ we must have $\mathbf\pi\cdot\mathbf x=0$. Applying this to the three known points generates a system of three linear equations in the components of $\mathbf\pi$, from which we see that $\mathbf\pi$ is a null vector of the matrix $$\begin{bmatrix}3&-4&2&1\\2&1&-3&0\\4&7&2&0\end{bmatrix}.$$ A straightforward row-reduction gives $[23:-16:10:-153]$, which corresponds to the equation $$23x-16y+10z=153.$$
