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

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

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

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    $\begingroup$ use family of planes $\endgroup$ – maveric Nov 16 '18 at 18:09
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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$

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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.

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