Curvature and torsion conditions tied to an auxiliary surface

Under what condition does the center of curvature of the helix

lie on the same cylinder as the helix itself?

Prove that for the curve

the segment of the (non-principal!) normal from a point of the curve to the axis $Oz$ equals the radius of torsion.

Prove that if at the point $M_{0}$ the curvature $k$ of the curve $\Gamma$ is nonzero, then $k$ equals the curvature of the projection of $\Gamma$ onto its osculating plane at the point $M_{0}$.

Under what condition does the center of curvature of the helix $x = a \cos t, y = a \sin t, z = b t$ lie on the same cylinder as the helix itself?

Prove that for the curve $x = a \operatorname {ch} t \cos t, y = a \operatorname {ch} t \sin t, z = a t$, the segments of the normal from a point on the curve to the $z$-axis equal the reciprocal of the absolute value of the torsion of the curve: $1 /\left|\varkappa \right|$.