The plane denoted by $$P_1 : 4x + 7y + 4z + 81 = 0$$ is rotated through a right angle about its line of intersection with the plane $$P_2 : 5x + 3y + 10z = 25$$. If the plane in its new position be denoted by $P$, and the distance of this plane from the origin is $\sqrt{d}$ then find $d$.
I tried
The equation of line of intersection is
$$(4x + 7y + 4z + 81) + k(4x+7y+4z+81) = 0$$
On
Let $r$ be the line of intersection of $P_1$ and $P_2$.
Hints:
(1) $(4,7,4)$ is a normal vector of $P_1$.
(2) If $v$ is a direction of $r$, (do you know how to find such direction?), then $(a,b,c)=v\times (4,7,4)$ could be a normal vector of $P$ (why?).
(3) Find a point that belongs to $P_1$ and $r$ and it will belong to $P$, say $p=(x_0,y_0,z_0)$ (why?).
(4) An equation for $P$ is given by $ax+by+cz+f=0$, where $ax_0+by_0+cz_0+f=0$.
(5) Now you can calculate such $d$.
The equation written by the OP is of plane passing through both the given planes .
The dot product of this equation and of initial equation is zero .
By doing this we get k=$-1$