If the line $\bar{r}=(\hat{\imath}-2 \hat{\jmath}+3 \hat{k})+\lambda(2 \hat{\imath}+\hat{\jmath}+2 \hat{k})$…
If the line $\bar{r}=(\hat{\imath}-2 \hat{\jmath}+3 \hat{k})+\lambda(2 \hat{\imath}+\hat{\jmath}+2 \hat{k})$ is parallel to the plane
$\bar{r} \cdot(3 \hat{\imath}-2 \hat{\jmath}-m \hat{k})=5$, then value of $m$ is
$-2$
$-3$
2
3
Solution
$\overline{\mathrm{r}}=\overline{\mathrm{a}}+\lambda \overline{\mathrm{b}}$ is parallel to the plane $\overline{\mathrm{b}} \cdot \overline{\mathrm{n}}=0$
Here $\overline{\mathrm{b}} \cdot \overline{\mathrm{n}}=(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}}) \cdot(3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-\mathrm{m} \hat{\mathrm{k}})=0$
$2(3)+1(-2)+2(-m)=0 \Rightarrow 6-2-2 m=0 \Rightarrow m=2$