Find the equation of the plane that contains Find the equation of the plane that contains,
the lines: $$
\frac{x-2}{2} = \frac{y+4}{3} = \frac{2 - z}{5}
$$
and
$$\begin{align}
x &= 3 + 4t \\
y &= -4 +6t \\
z &= 5 -10t.
\end{align}
$$
I'm not exactly sure on how to tackle this problem. 
 A: The line $\frac{x-2}{2} = \frac{y+4}{3} = \frac{2-z}{5}\,$ can be written as 
$$x=2+2t\,,\quad y=-4+3t \,,\quad z=2-5t \,, $$ 
and the line 
$$x = 3 + 4t \,,y = -4 + 6t \,,z = 5 - 10t \,.$$
The two lines have the same direction, since $ v_1=(2,3,-5) $ and $ v_2 = (4, 6,-10)\,, $ where $v_1$ and $v_2$ are the direction vectors of the two lines.
One can get two points lie in the plane. Putting $t=0$ in the equations of the lines gives $p_1=(2,-4,2)$ and $p_2=(3,-4,5)\,,$ which lie in the plane.
Constructing the vector $ v_3=p_2-p_1$ gives $v_3=(1,0,3)\,.$ Now, we can find the normal to the plane by taking the cross product of $v_3$ and $v_1$ or $v_2$. 
$$ n = v_3 \times v_2 \,. $$
Once that done, the equation of the plane is given by
$$ n.(X-p_1)=0 \Rightarrow n.( x-2,y-4,z-2 )=0 \,.$$
A: Hints: To find a plane all you need is a point and a normal vector. You can get one point from one line.
The normal vector $n$ needs to be perpendicular on both lines, thus perpendicular on both their directions. How do you find a vector perpendicular on two given directions? 
A: Hint: Note that the two lines are are parallel - both with direction vector $v = (2, 3, -5)$. Now for an equation of a plane you need a point and normal vector. The point should be easy to find.
For the normal vector, pick two point on each of the two lines given, say points $A$ and $B$. Then (try to draw a picture) a normal vector will be $v\times \vec{AB}$.
