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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 $a =1, b=1$
$$17+5 + 19c =0
\ 19c = -22
\ c=-{22 over 9}$$

$$vec n_3 = [1,1,-{22over 19}]
\ equiv [19,19,-22]$$

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 $14x +17y -17z +9 = 0$

Another hint:

Equations of all planes passing through the intersection of planes 2x-3y-z+1 = 0 and 3x+5y-4z+2=0 have a shape

$alpha (2x-3y-z+1) + beta (3x+5y-4z+2)=0$

The alpha and beta constants are not at the same time equal to 0. Suppose that alpha $neq 0$ and divide:

$frac{beta}{alpha}=gamma quad Rightarrow quad 2x-3y-z+1 + gamma (3x+5y-4z+2)=0$

The plane passes through the point $(3,-1,2) Rightarrow gamma = 4$ and the equation of the search plane will be

$ x + 17 y - 17 z + 9 = 0$

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.