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?
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 $\theta = \frac{\pi}{3}$