Spiral curve not possible, 10 deg 50 min deviation

Proof

A symmetrical spiral-circular-spiral curve needs the deflection angle to accommodate both spirals plus a circular arc:

$$\Delta\ge 2\theta_s,\qquad \Delta_c=\Delta-2\theta_s\ge0$$

Spiral angle of each transition curve:

$$\theta_s=\frac{L_s}{2R}=\frac{120}{2(400)}=0.15\ \text{rad}=8.594^\circ$$ $$2\theta_s=17.189^\circ$$

Deflection angle $\Delta=10^\circ50'=10.833^\circ$.

$$\Delta_c=\Delta-2\theta_s=10.833^\circ-17.189^\circ=-6.355^\circ<0$$

The central angle of the circular arc would be negative, which is impossible. The two spirals alone already turn through 17.19°, more than the available deflection of 10.83°, so the 120 m spiral cannot be introduced. Hence proved.

Possible solutions

1. Shorten the spiral. For the spirals to just meet ($\Delta_c=0$) with $R=400$ m: $L_s=R\Delta=75.6$ m. A shorter spiral (say 40 to 60 m) with a short circular arc is possible, but it must still satisfy the comfort and superelevation-run-off requirements.
2. Increase the radius. For $L_s=120$ m and $\Delta=10.83^\circ$, the radius must be at least $R=\dfrac{L_s}{\Delta}=\dfrac{120}{0.1891}=635$ m (pure spiral-spiral curve), and more to leave a circular arc. A larger radius also reduces the need for a long transition.
3. Use a pure (spiral-spiral) curve with $\theta_s=\Delta/2$ and no circular portion.
4. Use a simple circular curve with superelevation run-off provided on the straight. For a small deflection angle with a large radius the shift is tiny: here $s=\dfrac{L_s^2}{24R}=1.50$ m for 120 m. IRC does not require a transition for very large radii.
5. Shift the alignment (change the position of the intersection point) to get a larger deflection angle.