Let $\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{b}}=2 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}$. If
$\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{r}}, \overrightarrow{\mathrm{r}} \cdot(\alpha \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})=3$ and
$\overrightarrow{\mathrm{r}} \cdot(2 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}-\alpha \hat{\mathrm{k}})=-1, \alpha \in \mathrm{R}$, then the value of $\alpha+|\vec{r}|^{2}$ is equal to :
Correct Option: , 2
$\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{r}} \Rightarrow \overrightarrow{\mathrm{r}} \times(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}})=0$
$\overrightarrow{\mathrm{r}}=\vec{\lambda}(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \Rightarrow \overrightarrow{\mathrm{r}}=\vec{\lambda}(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}+2 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+5 \hat{\mathrm{k}})$
$\overrightarrow{\mathrm{r}}=\vec{\lambda}(3 \hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}})$.......(1)
$\overrightarrow{\mathrm{r}} \cdot(\alpha \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})=3$
Put $\overrightarrow{\mathrm{r}}$ from (1) $\alpha \lambda=1$.......(2)
$\overrightarrow{\mathrm{r}} \cdot(2 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}-\alpha \hat{\mathrm{k}})=-1$
Put $\overrightarrow{\mathrm{r}}$ from (1) $2 \lambda \alpha-\lambda=1$......(3)
Solve (2) & (3)
$\alpha=1, \quad \lambda=1$
$\Rightarrow \quad \overrightarrow{\mathrm{r}}=3 \hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}}$
$|\vec{r}|^{2}=14 \quad \& \quad \alpha=1$
$\alpha+|\vec{r}|^{2}=15$
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