System of Equations make a Trig Equation?

Let $A=3x+2y-z=4$ and $B=2x+y-3z=1$ . Because the planes meet at the same line, the plane governed by the 3rd equation is a linear sum of Equations A and B. Multiplying $A$ by $m$ , $B$ by $n$ , and adding the equations yields $(3m+2n)x+(2m+n)y+(-m-3n)z=(4m+n)=k$ . We can set $3m+2n=\sin{θ}$ , $2m+n=\cos{θ}$ , and $m+3n=2$ . This means that $m=2-3n$ , so $6-7n=\sin{θ}$ and $4-5n=\cos{θ}$ . By the Pythagorean Identity, $(6-7n)^2+(4-5n)^2=1$ . Solving this yields $2$ real values for $n$ , and by substitution, $2$ real values for $m$ . Using Vieta's formulas, the sum of the $2$ values of $4m+n$ comes out to be $\frac{-90}{37}$ and $2(90+37)=254$ which is the final answer.

Let $z = t$ .

Then $3x + 2y - t = 4 \Longrightarrow 6x + 4y - 2t = 8$

and $2x + y - 3t = 1 \Longrightarrow 6x + 3y - 9t = 3$ .

Subtracting these two equations give $y + 7t = 5 \Longrightarrow y = 5 - 7t$

Substituting $y = 5 - 7t$ into $3x + 2y - t = 4$ gives $3x + 2(5 - 7t) - t = 4 \Longrightarrow x = -2 + 5t$ .

Substituting $x = -2 + 5t$ , $y = 5 - 7t$ , and $z = t$ into $x \sin \theta + y \cos \theta - 2z = k$ gives $(-2 + 5t) \sin \theta + (5 - 7t) \cos \theta - 2t = k$ , which rearranges to $k = -2 \sin \theta + 5 \cos \theta + (5 \sin \theta - 7 \cos \theta - 2)t$ .

Substituting $\cos \theta = \pm \sqrt{1 - \sin^2 \theta}$ gives $k = -2 \sin \theta \pm 5\sqrt{1 - \sin^2 \theta} + (5 \sin \theta \mp \sqrt{1 - \sin^2 \theta} - 2)t$

For $k$ to be a constant, the coefficient of $t$ must be negative, so $5 \sin \theta \mp \sqrt{1 - \sin^2 \theta} - 2 = 0$ , which solves to $\sin \theta = \cfrac{5\sqrt{2} \pm 7 \sqrt{35}}{37\sqrt{2}}$ .

Substituting $\sin \theta = \cfrac{5\sqrt{2} \pm 7 \sqrt{35}}{37\sqrt{2}}$ into $k = -2 \sin \theta \pm 5\sqrt{1 - \sin^2 \theta}$ gives $k = \cfrac{-90 + 11\sqrt{70}}{74}$ and $k = \cfrac{-90 - 11\sqrt{70}}{74}$ , which has a sum of $-\cfrac{90}{37}$ .

Therefore, $a = 90$ , $b = 37$ , and $2a + 2b = \boxed{254}$ .