Let $\overrightarrow{\mathrm{a}}=2 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{b}}=7 \hat{\mathrm{i}}+\hat{\mathrm{j}}-6 \hat{\mathrm{k}}$.
If $\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{r}} \cdot(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})=-3$, then $\overrightarrow{\mathrm{r}} \cdot(2 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+\hat{\mathrm{k}})$
is equal to :
Correct Option: 1
$\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{b}}=0$
$\Rightarrow \overrightarrow{\mathbf{r}} \times(\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}})=0$
$\Rightarrow \overrightarrow{\mathrm{r}}=\lambda(\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}})$
$\Rightarrow \overrightarrow{\mathrm{r}}=\lambda(-5 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+10 \hat{\mathrm{k}})$
Also $\overrightarrow{\mathrm{r}} \cdot(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})=-3$
$\Rightarrow \lambda(-5-8+10)=-3$
$\lambda=1$
Now $\overrightarrow{\mathrm{r}}=-5 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+10 \hat{\mathrm{k}}$
$=\overrightarrow{\mathrm{r}} \cdot(2 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+\hat{\mathrm{k}})$
$=-10+12+10=12$
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