Mathematics - 3D-Coordinate Geometry Question with Solution | TestHub

We have,

$\overrightarrow{r}·\left(\mathrm{\lambda }\hat{\mathrm{i}}+5\hat{\mathrm{j}}-\hat{\mathrm{k}}\right)=5$

$\overrightarrow{r}·\left(2\hat{\mathrm{i}}-\mathrm{\lambda }\hat{\mathrm{j}}+3\hat{\mathrm{k}}\right)=0$

We know that two planes $\overrightarrow{r}·{\overrightarrow{n}}_1=d_1$ and $\overrightarrow{r}·{\overrightarrow{n}}_2=d_2$ are perpendicular if

${\overrightarrow{n}}_1·{\overrightarrow{n}}_2=0$

Therefore,

$\left(\mathrm{\lambda }\hat{\mathrm{i}}+5\hat{\mathrm{j}}-\hat{\mathrm{k}}\right)·\left(2\hat{\mathrm{i}}-\mathrm{\lambda }\hat{\mathrm{j}}+3\hat{\mathrm{k}}\right)=0$

$\Rightarrow 2\lambda -5\lambda -3=0$

$\Rightarrow -3\lambda -3=0$

$\Rightarrow \lambda =-1$

Therefore, ${\lambda }^2+\lambda =0$