Determine the scalarform of a plane that contains the line $L$ and goes through the midpoint $P$ So I need to determine the scalarform of a plane that contain $L:\begin{cases} 3x + 4y + z = 5 \\
x − y = −6 \end{cases}$ and goes through the midpoint of $(1, 1, 2)$ and $(3, 1, 4)$.
My try solving this. Call the midpoint $P$ then $P:(2,1,3)$. The direction vector $v$ of the line is obtained by the crossproduct of the normal of each plane, so $(3,4,1)\times (1,-1,0)=(1,1,-7)$. Thus we have $x+y-7z=d$ and putting in $(2,1,3)$ gives $x+y-7z=20$. But the answer is $13x + 36y + 7z = 83$.
 A: Your approach is not correct. $x+y−7z=-18$ is the plane which is orthogonal to the line $L$ and passes through $(2,1,3)$.
Notice that all the planes which contain the line $L$ (sheaf of planes) are given by
$$a(3x+4y+z-5)+b(x-y+6)=0\quad \text{with $a,b\in\mathbb{R}$}.$$
We have to find $a,b$ such that the above equation is satisfied for $(x,y,z)=(2,1,3)$:
$$a(6+4+3-5)+b(2-1+6)=0\implies 8a+7b=0.$$
Therefore may take $a=7$ and $b=-8$:
$$7(3x+4y+z-5)-8(x-y+6)=13x + 36y + 7z - 83=0$$
and your answer is confirmed.
A: $(1, 1, -7)$ is the direction vector of the line of intersection of the two given planes (line $L$).  It is not the normal vector of the required plane.
In addition to the direction vector of line $L$, you need a point on that line.  Such a point can be found by Gauss-Jordan elimination of the two linear equations of the two planes
$ 3 x + 4 y + z = 5 \hspace{25pt}(1) $
$ x - y = -6 \hspace{25pt}(2)$
Set $x = 0$, and solve the resulting system for $y$ and $z$
$ 4 y + z = 5 \hspace{25pt}(1')$
$ - y = - 6 \hspace{25pt}(2')$
So $(2')$ implies $ y = 6 $ and then $(1')$ implies $z = -19 $
So a point on line $L$ is $Q=(0, 6, -19) $
The midpoint you found to be $P = (2,1,3)$
Calculate the vector
$QP = P - Q = (2, -5, 22) $
And you have the direction vector of line $L$, namely,
$ V = (1, 1, -7)$
Hence, the normal to the plane is
$ N = QP \times V = (2, -5, 22) \times (1, 1, -7) = (13, 36, 7 ) $
The equation of the plane is
$ N \cdot (p - Q) = 0 $
$ (13, 36, 7) \cdot ((x,y,z) - (0, 6, -19) ) = 0 $
which simplifies to
$ 13 z + 36 y + 7 z - 83 = 0 $
