Equation of a plane given point and line 
Find the equation of the plane containing the point $(1, 3,−2)$ and
  the line $x = 3 + t$, $y = −2 + 4t$,  $z = 1 − 2t$.

Don't know where to start?
 A: Hint:By taking $t=1$ which can be any other real number into the equation of the line we have: $$P:(4,2,-1)$$ as a point lying on the line. Now write the equation of the line passing $P$ and the point you have above $A:(1,3,-2)$. Then find the cross product of two leading vector, one is for the first line, and the other is for the line $PA$. Use it for the normal vector for the desired plane. 
A: First, lets find the point lying on the line when $t=0$:
$\begin{cases} x = 3 + 0 \\ y = −2 + 0 \\ z = 1 − 0 \end{cases}$ Thus we have $\rightarrow$ $A(3,-2,1)$
Then, find another point, we denote as B, when $t = 1$:
$\begin{cases} x = 3 + 1 \\ y = −2 + 4 \\ z = 1 − 2 \end{cases}$ Thus we have $\rightarrow$ $B(4,2,-1)$
Then determine the vectors on the plane, then obtain the normal vector by taking the cross product which will finally give you the desired equation. If you need more help just ask. 
A: point A(1,3,-2) act as origin 
point B which is on the plane has point(3,-2,1)
from three equations we can get vector of a plane <1,4,-2>
then point AB vector is B-A =(2,-5,0)
to get normal vector we multiply AB*V(vector)
{2 -5 0}    =(10 4 13)                                               
{1 4 -2}

10(X-1)+4(Y-3)+13(Z-2)
10X+4Y+13Z=48
