Plane and solid analytic geometry, by William F. Osgood and William C. Graustein.

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ANALYTIC GEOMETRY

Second Method. Let N be any normal to the required plane.
Since the plane contains the two given points, N is perpendicular to the line L1 joining these points. Since the plane is
perpendicular to the given plane (2), N is perpendicular to
any line L2 normal to (2). Thus N is a common perpendicular
to the lines L1 and L2.
The direction components of L1 are, by Ch. XVIII, ~ 4,
3-(-6), -2-0, 9-(-4),i.e.
9, -2, 13;
those of L2 are 2, -1, 4.
Consequently, by (6), Ch. XVIII, ~ 5, the direction components
of N are
-2 13 13 9 9 - 2
-1 4 42 2 -1
i.e. 5, -10, - 5, or 1, - 2, - 1.
Our problem is now reduced to that of finding the equation
of the plane which passes through one of the given points,
say (3, - 2, 9), and has 1, -2, -1 as the direction components of its normals. This equation is
(x - 3)- 2(y +2)-1 (z -9)=0,
or x-2y-z + 2 =0.
Example 3. Find the equation of the plane passing through
the point (2, 5, - 8) and perpendicular to each of the
planes:
2x-3y+4z+1=0, 4x+y-2z+6=0.
Either of the methods employed in the previous example is
applicable. We choose the latter. A normal N to the required plane is perpendicular to the normals to both the given
planes. These have, respectively, the direction components
2, -3, 4 and 4, 1, - 2. Consequently, the direction components of N are 2, 20, 14 or 1, 10, 7.
The equation of the plane through (2, 5, - 8) with 1, 10, 7
as the direction components of its normals is