<s>[INST] Is there an elegant way to find the equation of a plane given its line and angle of intersection with another plane? This is a problem I give sometimes to my calculus students:

Let $\mathcal{Q}$ be the plane defined by $x+2y+3z=6.$ Find a plane $\mathcal{P}$ that meets $\mathcal{Q}$ at an angle of $\theta = \frac{\pi}{3}$ such that $\mathcal{P},\mathcal{Q}$ intersect along the line $\langle 3,0,1 \rangle+t \langle -3,1,\frac{1}{3} \rangle.$ (Hint: Write $\mathcal{P}$ as $ax+by+cz=6$, and use the conditions to solve for $a,b,c.$)


The intended solution: We know that $\mathcal{P}$ contains the point $\langle 3,0,1\rangle$ and also $\langle 3,0,1 \rangle+t \langle -3,1,\frac{1}{3} \rangle$ for any $t$, in particular for $t=1.$ Then we have:
$$
\langle 3,0,1\rangle\cdot \langle a,b,c\rangle = 6,\qquad \langle 0,1,4/3\rangle\cdot \langle a,b,c\rangle = 6
$$
We also know the angle of intersection:
$$
\frac{\langle a,b,c\rangle \cdot \langle 1,2,3 \rangle}{\sqrt{a^2+b^2+c^2}\;\sqrt{14}}=\cos(\pi/3)=\frac{1}{2}
$$Putting these together, we are solving:
$$
\begin{cases}
3a+c=6\\
b+\frac{4}{3}c = 6\\
\frac{a+2b+3c}{\sqrt{14(a^2+b^2+c^2)}}=\frac{1}{2}
\end{cases}
$$Solving the first two equations for $a$ gives $c=6-3a$, $b=4a-2.$ Plugging this into the third equation gives:
$$
56=a^2+(4a-2)^2+(6-3a)^2
$$
This is just a quadratic equation in $a.$ Solving it produces
$$
\{a,b,c\}=\left\{1\mp\frac{\sqrt{273}}{13},2\mp\frac{4\sqrt{273}}{13},3\pm\frac{3\sqrt{273}}{13} \right\}
$$
Not terrible but not entirely straightforward. I'm wondering if there is a simpler way to reach this conclusion, maybe by exploiting the normal vectors of the planes?
 [/INST] a direction within the plane $Q$  that is orthogonal to the line is
$(1,2,3)$ crossed with $(-9,3,1);$ this is $7(-1,-4,3).$  Let us name $u= (-1,-4,3).$
Length of $u$ is $\sqrt{26},$   so make a unit vector
$$  v = \frac{1}{\sqrt{26}}(-1,-4,3) $$
The unit normal to $Q$ is
$$  w = \frac{1}{\sqrt{14}} (1,2,3)  $$
Next let $$  N = v \cos \theta + w \sin \theta  $$
be a unit vector at angle $\theta$ from $v.$  You want $\theta = \frac{\pi}{3}$
</s>