Given ell_1: (x-2)/(λ)=tfracy-11=tfracz-20 and ell_2…

Given $\ell_1:\ \tfrac{x-2}{\lambda}=\tfrac{y-1}1=\tfrac{z-2}0$ and $\ell_2:\ \tfrac{x-5}2=\tfrac{y-2}3=\tfrac{z-3}1$: $1^\circ$ determine $\lambda$ so that they intersect; $2^\circ$ find the intersection point; $3^\circ$ write the line through $O$ perpendicular to the plane of the two lines.

Explanation of Method

Choosing λ so that two lines intersect, then finding the normal through O to their plane.

Step 1: $1^\circ$ Coplanarity

\[ \begin{vmatrix}3&1&1\\\lambda&1&0\\2&3&1\end{vmatrix}=3-\lambda+(3\lambda-2)=2\lambda+1=0 . \]

\[\lambda=-\tfrac12\]

Step 2: $2^\circ$ Intersection

$\tfrac{z-2}0$ means $z=2$ on $\ell_1$. On $\ell_2$, $z=3+s=2$ gives $s=-1$, so the point is $(3,-1,2)$. On $\ell_1$, $t=-2$ gives $x=2+1=3$ and $y=1-2=-1$. $\checkmark$

\[P(3,-1,2)\]

Step 3: $3^\circ$ Normal through $O$

\[ \vec n=\{-1,2,0\}\times\{2,3,1\}=\{2,1,-7\},\qquad \text{plane: }2x+y-7z+9=0 . \]

\[\frac x2=\frac y1=\frac z{-7}\]

Check: $M_1(2,1,2)$, $M_2(5,2,3)$ and $P$ all satisfy $2x+y-7z+9=0$.

$P(3,-1,2)$ $O$ $\ell_1$ $\ell_2$ $\tfrac x2=\tfrac y1=\tfrac z{-7}$

For $\lambda=-\tfrac12$ the lines meet at $P(3,-1,2)$ and span the plane $2x+y-7z+9=0$ (gray); the teal line through $O$ is perpendicular to it.

Result

\[\boxed{\,\lambda=-\tfrac12\,}\]

\[\boxed{\,P(3,-1,2)\,}\]

\[\boxed{\,\frac x2=\frac y1=\frac z{-7}\,}\]