The equation of a plane containing the lines $\overline{\mathrm{r}}=(\hat{\imath}+2 \hat{\jmath}-4…
The equation of a plane containing the lines $\overline{\mathrm{r}}=(\hat{\imath}+2 \hat{\jmath}-4 \hat{\mathrm{k}})+\lambda(2 \hat{\imath}+3 \hat{\jmath}+6 \hat{\mathrm{k}})$ and
$\overline{\mathbf{r}}=(\hat{\imath}+3 \hat{\jmath}+4 \hat{\mathrm{k}})+\mu(\hat{\imath}+\hat{\jmath}-\hat{\mathrm{k}})$ is
$9 x+8 y+z+11=0$
$9 x-8 y-z-11=0$
$9 x-8 y-z+11=0$
$9 x-8 y+z+11=0$
Solution
Normal vector of a plane would be perpendicular to both the given lines and parallel to their cross product.
Now $\bar{\ell}_{1} \times \bar{\ell}_{2}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 6 \\ 1 & 1 & -1\end{array}\right|$
$=-9 \hat{\mathrm{i}}+8 \hat{\mathrm{j}}-\hat{\mathrm{k}}$ i.e. $\Rightarrow 9,-8,1$ are d.r. of normal to a plane
Let $\bar{a}=\hat{i}+3 \hat{j}+4 \hat{k}$ and $\bar{n}=9 \hat{i}-8 \hat{j}+\hat{k}$
$\therefore \overline{\mathrm{r}} \cdot(9 \hat{\mathrm{i}}-8 \hat{\mathrm{j}}+\hat{\mathrm{k}}) \quad=9(1)+(-8)(3)+4 \times 1=9-24+4$
$\overline{\mathrm{r}} \cdot(9 \hat{\mathrm{i}}-8 \hat{\mathrm{j}}+\hat{\mathrm{k}}) \quad=-11 \Rightarrow 9 \mathrm{x}-8 \mathrm{y}+\mathrm{z}+11=0$ is the equation of plane.