In each case, determine whether or not the lines have a single point of intersection. If they do, give an equation of a plane containing them. a) r₁ = (5t, 2t 1,3t 3) and r₂ = (t−6, -t + 5, t - 9)

$\left(a\right)$ Given that ${\overrightarrow{r}}_1=〈5t,2t-1,3t-3〉$ and $\overrightarrow{r_2}=〈t-6,-t+5,t-9〉$.

We have to find the equation of the plane.

Now, ${\overrightarrow{r}}_1\text{ at }{t}_1=〈5t_1,2t_1-1,3t_1-3〉$

And $\overrightarrow{r_2}\text{ at }{t}_2=〈t_2-6,-t_2+5,t_2-9〉$

For intersection point

$\begin{array}{c}5t_1=t_2-\mathrm{6........}\left(1\right)\\ 2t_1-1=-t_2+\mathrm{5......}\left(2\right)\\ 3t_1-3=t_2-\mathrm{9........}\left(3\right)\end{array}$

Now, solve equation $\left(1\right), \left(2\right) and \left(3\right)$, we get

$t_1=0\text{ and }{t}_2=6$

If we substitute $t_1=0\text{ and }{t}_2=6$ in equation $\left(3\right)$, it satisfy the third equation.

Hence, ${\overrightarrow{r}}_1$ and $\overrightarrow{r_2}$ intersects each other.

Now at $t_1=0$, ${\overrightarrow{r}}_1=〈0,-1,-3〉$.