= Solution
Choose the helix handedness so its unit tangent is $\mathbf t=\cos\alpha\,\mathbf i+\sin\alpha\,\mathbf e_\phi$. Let $q=\Omega_0-\Omega$; the body rotates at $+\Omega\mathbf i$, while the flagellum rotates at $-q\mathbf i$ in the laboratory, since its relative motor rotation is $-\Omega_0\mathbf i$. Both translate at $-U\mathbf i$. Thus the relative fluid <velocity> on a flagellar element is $\mathbf w=U\mathbf i+bq\mathbf e_\phi$.
Low-Reynolds-number <resistive-force theory> gives <force> per unit length on the flagellum
$$
\mathbf f=K_N\mathbf w+(K_T-K_N)(\mathbf w\cdot\mathbf t)\mathbf t.
$$
Its axial and azimuthal components are
$$
f_x=(K_N\sin^2\alpha+K_T\cos^2\alpha)U-(K_N-K_T)bq\sin\alpha\cos\alpha,
$$
$$
f_\phi=(K_N\cos^2\alpha+K_T\sin^2\alpha)bq-(K_N-K_T)U\sin\alpha\cos\alpha.
$$
Define the integrated <axial resistance matrix of a slender helix> coefficients
$$
A=L(K_N\sin^2\alpha+K_T\cos^2\alpha),\quad
C=bL(K_N-K_T)\sin\alpha\cos\alpha,\quad
H=b^2L(K_N\cos^2\alpha+K_T\sin^2\alpha).
$$
The sphere has translational resistance $A_b=6\pi\mu a$ by the <Stokes drag law> and rotational resistance $H_b=8\pi\mu a^3$. Its hydrodynamic <force> is $+A_bU\mathbf i$ and its <torque> $-H_b\Omega\mathbf i$. Integration of $f_x$ and $bf_\phi$ gives the zero total <force> and <torque> conditions
$$
\boxed{(A+A_b)U=Cq,\qquad CU=Hq-H_b\Omega.}
$$
Dividing the first by $L$ and expanding the coefficients gives exactly the two requested balances. The internal motor <torque> cancels in the total <torque> balance; it is not an external applied <torque>.
Eliminate $U$ and use $q+\Omega=\Omega_0$. The effective flagellar rotational resistance is $H_e=H-C^2/(A+A_b)$, so
$$
q=\frac{H_b\Omega_0}{H_b+H_e},\qquad
\boxed{\Omega=\frac{H_e\Omega_0}{H_b+H_e},\qquad
U=\frac{CH_b\Omega_0}{(A+A_b)(H+H_b)-C^2}.}
$$
The <determinant of helical resistance in resistive-force theory> is $AH-C^2=b^2L^2K_NK_T>0$, hence the denominator is positive. Opposite handedness changes the sign of the translation-rotation coupling and reverses the swimming direction.
For $K_T=K$, $K_N=2K$, $\alpha=\pi/4$ and $a=b$, the coefficients are $A=3KL/2$, $C=KaL/2$, $H=3Ka^2L/2$. Therefore
$$
\boxed{U=\frac{4\pi\mu KL a^2\Omega_0}{2K^2L^2+21\pi\mu aKL+48\pi^2\mu^2a^2}.}
$$
Equivalently, with $h=KL/(\pi\mu a)$, $U=a\Omega_0\,4h/(2h^2+21h+48)$. These are positive speed magnitudes for the selected handedness; the translational <velocity> is $-U\mathbf i$.
This <helical microswimmer with a spherical head> approximation neglects head-tail hydrodynamic interactions as permitted, treats the flagellum as slender and rigid, and averages away transverse <forces> associated with finite helical ends. The flow is quasistatic <Stokes flow>, so neither body inertia nor fluid inertia enters the balance.
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