Find the cylindrical surface through the curve…
Find the cylindrical surface through the curve $(x-1)^2+(y+3)^2+(z-2)^2=25,\qquad x+y-z+2=0$ with generators parallel to $1^\circ$ the $x$-axis and $2^\circ$ the line $x=y,\ z=c$.
Explanation of Method
Cylinders through the circle cut from a sphere by a plane, with two different generator directions.
Step 1: $1^\circ$ Direction $\{1,0,0\}$
The generator through $M$ meets the plane at $(z-y-2,\ y,\ z)$. This point must lie on the sphere: $(z-y-3)^2+(y+3)^2+(z-2)^2=25$.
\[2y^2+2z^2-2yz+12y-10z-3=0\]
Step 2: $2^\circ$ Direction $\{1,1,0\}$
With $t=\tfrac{z-x-y-2}{2}$, the trace is $X_0=\tfrac{x-y+z-2}2$, $Y_0=\tfrac{-x+y+z-2}2$, $Z_0=z$. Substituting into the sphere and multiplying by $4$, with $u=x-y$: $(u+z-4)^2+(-u+z+4)^2+4(z-2)^2=100$.
Expanding term by term: $(u+z-4)^2=u^2+z^2+16+2uz-8u-8z$.
\[ (-u+z+4)^2=u^2+z^2+16-2uz-8u+8z . \]
Their sum is $2u^2+2z^2+32-16u$. Then
\[ 2u^2-16u+6z^2-16z-52=0\iff u^2-8u+3z^2-8z-26=0 . \]
\[(x-y)^2-8(x-y)+3z^2-8z-26=0\]
Check with a point of the curve: $x=1$ gives $y=-2+s,\ z=1+s$, where $s=\tfrac{\sqrt{46}}2$ ($s^2=\tfrac{23}2$). Then $u=3-s$, and
\[ (20.5-6s)-(24-8s)+(37.5+6s)-(8+8s)-26=0.\ \checkmark \]
Result
\[\boxed{\,2y^2+2z^2-2yz+12y-10z-3=0\,}\]
\[\boxed{\,(x-y)^2-8(x-y)+3z^2-8z-26=0\,}\]
Remark
Correction: the source expands $(-u+z+4)^2$ with $+8u$ instead of $-8u$, so the linear terms cancel and it loses $-8(x-y)$. Its answer $(x-y)^2+3z^2-8z-26=0$ fails at the point above (it gives $24-8s\approx-3.13$).
A quick sanity test: the circle's center is $K\left(\tfrac53,-\tfrac73,\tfrac43\right)$, with $x-y=4$. The cylinder must be symmetric about $x-y=4$, not about $x-y=0$.