Find the equation of the plane that passes through the line of intersection of the planes $2x-3y-z +1 =0$ and $3x+5y-4z+2=0$, and that also passes through the point $(3,-1,2)$
$\vec n_1 = [2,-3,-1]$
$\vec n_2 = [3,5,-4]$
$\vec n_1 \times \vec n_2 = [17,5,19]$
$[17,5,19]$ is the direction vector of the line of intersection.
now,
$$[17,5,19] \cdot \vec n_3 = 0
\\ [17,5,19] \cdot [a,b,c] =0
\\ 17a+5b+19c = 0$$
Let $a =1, b=1$
$$17+5 + 19c =0
\\ 19c = -22
\\ c=-{22 \over 9}$$
$$\vec n_3 = [1,1,-{22\over 19}]
\\ \equiv [19,19,-22]$$
So the scalar equation is $19x +19y- 22z+ D= 0$
Substitute $(3,-1,2)$
$$19(3) +19(-1) -22(2) +D = 0
\\ D = 6$$
$19x -19y -22z + 6 =0$ is the equation of the plane.
However the answer says it is $14x +17y -17z +9 = 0$
Best Answer
Find two points $P$ and $Q$ on the line of intersection of the planes $2x−3y−z+1=0$ and $3x+5y−4z+2=0.$
Then with $A=(3,−1,2)$ construct vectors $AP$ and $AQ$
The normal vector to your plane is the common perpendicular of $AP$ and $AQ$
Having a point and the normal vector you can easily find the equation of the plane.