Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

$$\begin{array}{rl}& \overline{\mathrm{r}}=(\hat{\mathrm{i}}-\hat{\mathrm{j}})+\lambda (\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})+\mu (\hat{\mathrm{i}}-2\hat{\mathrm{j}}+3\hat{\mathrm{k}})\\ & \overline{\mathrm{a}}=\hat{\mathrm{i}}-\hat{\mathrm{j}},\overline{\mathrm{A}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}},\overline{\mathrm{B}}=\hat{\mathrm{i}}-2\hat{\mathrm{j}}+3\hat{\mathrm{k}}\end{array}$$ $\overline{\mathrm{n}}$ is $\perp$ er $\overline{\mathrm{A}}$ and $\overline{\mathrm{B}}$ $$\begin{array}{rl}& \therefore \overline{\mathrm{n}}=\overline{\mathrm{A}}\times \overline{\mathrm{B}}=\left|\begin{array}{ccc}\hat{\mathrm{i}}& \hat{\mathrm{j}}& \hat{\mathrm{k}}\\ 1& 1& 1\\ 1& -2& 3\end{array}\right|\\ & =\hat{\mathrm{i}}(3+2)-\hat{\mathrm{j}}(3-1)+\hat{\mathrm{k}}(-2-1)=5\hat{\mathrm{i}}-2\hat{\mathrm{j}}-3\hat{\mathrm{k}}\\ & \overline{\mathrm{a}}\cdot \overline{\mathrm{n}}=\mathrm{d}=(\hat{\mathrm{i}}-\hat{\mathrm{j}})\cdot (5\hat{\mathrm{i}}-2\hat{\mathrm{j}}-3\hat{\mathrm{k}})=5+2=7\\ & \therefore \overline{\mathrm{r}}\cdot (5\hat{\mathrm{i}}-2\hat{\mathrm{j}}-3\hat{\mathrm{k}})=7\\ & \Rightarrow \text{ Cartesian equation is }5\mathrm{x}-2\mathrm{y}-3\mathrm{z}-7=0\end{array}$$