Finding the two planes which contains a given line and forms an angle of $60^\circ$ with a given plane I am given the following question:

Find the equations of the two planes that contain the line $r \{ x=z+1=y+2$ and form a $60^\circ$ angle with the plane $ \pi \{ x+2y-3z+2=0$

My solution:
Lets call the normal vector of the planes wanted $\vec{n}=(a,b,c)$
The vector of the line is $\vec{v} =(1,1,1)$ and since it is perpendicular to the planes we want,
$$
(1,1,1) \cdot (a,b,c) = 0 \therefore a+b+c = 0
$$
Also, since the angles between the plane $\pi$ and our new planes is $60^\circ$ we have
$$
\frac{1}{2}=\frac{\vert (a,b,c) \cdot (1,2,-3) \vert}{\sqrt{a^2+b^2+c^2} \sqrt{14}}
$$
I need another equation to finish the exercise, but I'm not sure what is that third equation.
Textbook's answer
$$
\pi_1 \{ 2x-3y+z-5=0\\
\pi_2 \{ 3x -y-2z-4=0
$$
Thank you.

Since as has been oft remarked the lenght of $(a,b,c)$ is variable let us chose the length to be $\sqrt{14}$ in order to cancel with the other $\sqrt{14}$
So we have the equations:
$$a+b+c=0$$
$$a+2b-3c=\frac{1}{2}\sqrt{14}\sqrt{a^2+b^2+c^2}=\frac{1}{2}14=7$$
$$a^2+b^2+c^2=14$$
If we take the first two 
$$a+b+c=0$$
$$a+2b-3c=7$$
and solve them we get
$$a=-5c-7$$
$$b=4c+7$$
If we now substitute into the third equation and simplify we get 
$$c^2+3c+2=0$$ so $c=-1,-2$.
This gives us the two vectors
$$(-2,3,-1) \text{and} (3,-1,-2).$$
And the two equations
$$2x-3y+z=d$$
$$3x-y-2z=d$$
The value of $d$ can then be calculated in each case since the plane passes through $(0,-2,-1)$. 
