Given two equation of lines in $3D$

Question

Plane $P$ contains the lines

\begin{align}L_1:&\frac{x-1}{k}=\frac{y-2}{2}=\frac{z-3}{1}=3\\ L_2:&\frac{x-2}{2}=\frac{y-3}{k}=\frac{z-4}{3}\end{align}

Prove that $L_1$ and $L_2$ intersect and the equation of plane is $x-y+1=0$.

I got the answer the answer of first part by susbtituting parametric equation of points in other equation of line .

But I am stuck how to find the plane

Answer 1

\begin{align}L_1:\frac{x-1}{k}=\frac{y-2}{2}=\frac{z-3}{3}=\lambda\\ L_2:\frac{x-2}{2}=\frac{y-3}{k}=\frac{z-4}{3}=\epsilon\end{align}

Vector parallel to $L_1:\vec{r_1}=k\underline{i}+2 \underline{j}+3\underline{k}$

Vector parallel to $L_2:\vec{r_2}=2\underline{i}+k\underline{j}+3\underline{k}$

Normal vector to required plane: $\vec{n}=\vec{r_1}\times \vec{r_2}$

Point on plane will be $(1,2,3)$ as it lies on line $L_1$

So equation of required plane is : $$(\vec{r}-\underline{i}-2\underline{j}-3\underline{k})\cdot \vec{n}=0$$

Where $\vec{r}=x\underline{i}+y\underline{j}+3\underline{k}$

Answer 2

As you already did first part I'll tell about second part. $\frac{x-2}{2}=\frac{y-3}{k}=\frac{z-4}{3}=a$ where a is any variable. Since you can find value of k yourself so just subtitue it in the above equation I wrote. Now you can get point let name it $P_2$ $(2a+2, ka+3, 3a+4)$ and from $L_1$ you can get $P_1$ the find direction ratios and now you're ready to form equation of plane.