If lines $\frac{\mathrm{x}-2}{1}=\frac{\mathrm{y}-4}{4}=\frac{\mathrm{z}-6}{7}\text{ and }\frac{(\mathrm{x}+1)}{3}=\frac{\mathrm{y}+3}{5}=\frac{(\mathrm{z}+5)}{\mathrm{r}}$ are coplanar, then find the value of r.

Given lines are,
$\frac{\mathrm{x}-2}{1}=\frac{\mathrm{y}-4}{4}=\frac{\mathrm{z}-6}{7}\text{ and }\frac{\mathrm{x}+1}{3}=\frac{\mathrm{y}+3}{5}=\frac{\mathrm{z}+5}{\mathrm{r}}$
Here, the points and direction ratios of given lines are
$\mathrm{and} \begin{array}{r}{\mathrm{x}}_1=2,\mathrm{y}=4,{\mathrm{z}}_1=6\\ {\mathrm{a}}_1=1,{\mathrm{b}}_1=4,{\mathrm{c}}_1=7\\ {\mathrm{x}}_2=-1,{\mathrm{y}}_2=-3,{\mathrm{z}}_2=-5\\ {\mathrm{a}}_2=3,{\mathrm{b}}_2=5,{\mathrm{c}}_{?}=\mathrm{r}\end{array}$
Now, these lines are coplanar, if
$\left|\begin{array}{ccc}{\mathrm{x}}_2-{\mathrm{x}}_1& {\mathrm{y}}_2-{\mathrm{y}}_1& {\mathrm{z}}_2-{\mathrm{z}}_1\\ {\mathrm{a}}_1& {\mathrm{b}}_1& {\mathrm{c}}_1\\ {\mathrm{a}}_2& {\mathrm{b}}_2& {\mathrm{c}}_2\end{array}\right|=0$
$\text{ Consider, }\left|\begin{array}{ccc}-1-2& -3-4& -5-6\\ 1& 4& 7\\ 3& 5& \mathrm{r}\end{array}\right|=\left|\begin{array}{r}-3\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}-7\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}-11\\ 1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}4\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}7\\ 3\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}5\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{r}\end{array}\right|$
$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\left|\begin{array}{rrr}-3& -7& -11\\ 1& 4& 7\\ 3& 5& \mathrm{r}\end{array}\right|\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}-3(4\mathrm{r}-35)+7(\mathrm{r}-21)-11(5-12)\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}-12\mathrm{r}+(-35\times -3)+7\mathrm{r}-147-55+132\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}-12\mathrm{r}-105+7\mathrm{r}-147-55+132\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}-5\mathrm{r}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=-35\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{r}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=7\end{array}$