Through M_1(1,2,-1) draw a line intersecting ell_1…

Through $M_1(1,2,-1)$ draw a line intersecting $\ell_1:\ \tfrac{x-1}{2}=\tfrac{y+1}{6}=\tfrac{z+3}{3}$ and $\ell_2:\ \tfrac{x-2}{3}=\tfrac y1=\tfrac{z+3}{-1}$.

Step 1: Collinearity condition

Let $A=(1+2s,-1+6s,-3+3s)\in\ell_1$ and $B=(2+3t,t,-3-t)\in\ell_2$. The points $M_1,A,B$ must be collinear, i.e. $\overrightarrow{M_1A}\times\overrightarrow{M_1B}=\vec0$: \[2+5t-6s-9st=0,\quad 2+6t-7s-11st=0,\quad 3+9t-10s-16st=0 .\]

The first two give $t=\tfrac{s}{1-2s}$ and $3s^2-5s+2=0$, so $s=1$ or $s=\tfrac23$. Only $(s,t)=(1,-1)$ satisfies the third equation.

\[ A(3,5,0),\qquad B(-1,-1,-2),\qquad \overrightarrow{M_1A}=\{2,3,1\}=-\overrightarrow{M_1B}. \]

\[\frac{x-1}{2}=\frac{y-2}{3}=\frac{z+1}{1}\]

Step 2: Alternative: two planes

The required line is the intersection of the plane through $M_1$ and $\ell_1$ with the plane through $M_1$ and $\ell_2$: \[\vec n_1=\{0,3,2\}\times\{2,6,3\}=\{-3,4,-6\},\qquad \vec n_2=\{1,-2,-2\}\times\{3,1,-1\}=\{4,-5,7\}.\]

The direction is $\vec n_1\times\vec n_2=\{-2,-3,-1\}\parallel\{2,3,1\}$, the same line.