Rotation of a plane with respect to a straight line The plane x-2y+3z=0 is rotated through a right angle about its line of intersection with the plane 2x+3y-4z-5=0; Find the equation of the plane in its new position?
 A: You can first find what is the intersection of the first plane, of equation $x - 2y + 3z = 0$, with the second plane, given by $2x + 3y - 4z = 5$. This amounts to solving a linear system and the solution is a line. One way to represent this line, if I calculated correctly is
$$ r: t \in \mathbb{R} \mapsto \left(\frac{10}{7}, \frac{5}{7}, 0 \right) + t(-1, 10, 7) $$
Now, notice that the vector $n = (1, -2, 3)$ is orthogonal to the plane $x - 2y + 3z = 0$. The vector $v = (-1, 10, 7)$ that gives the direction of line $r$ must be orthogonal to $n$, clearly. Finally, if you rotate the first plane by $90$ degrees around $r$ you get a vector that is orthogonal to both $n$ and $v$. Taking
$$ w =  n \times v = (-44, -10, 8)$$
you get a vector that is normal to your rotated plane. Moreover, the plane still contains any point of line $r$, for instance $p = (10/7, 5/7, 0)$. There is only one plane with normal vector $w$ passing through $p$, and its equation is given by
$$ -44x - 10 y +8z = c$$
where $c = -44 \cdot \frac{10}{7} - 10 \cdot \frac{5}{7} + 8 \cdot 0 = -70$.
