Let $\overrightarrow a$ = $\widehat i$ + 2$\widehat j$ $-$ 3$\widehat k$ and $\overrightarrow b = 2\widehat i$ $-$ 3$\widehat j$ + 5$\widehat k$. If $\overrightarrow r$ $\times$ $\overrightarrow a$ = $\overrightarrow b$ $\times$ $\overrightarrow r$,

$\overrightarrow r$ . $\left( {\alpha \widehat i + 2\widehat j + \widehat k} \right)$ = 3 and $$\overrightarrow r \,.\,\left( {2\widehat i + 5\widehat j - \alpha \widehat k} \right)$$ = $-$1, $\alpha$ $\in$ R, then the

value of $\alpha$ + ${\left| {\overrightarrow r } \right|^2}$ is equal to :

Given $\overrightarrow r$ $\times$ $\overrightarrow a$ = $\overrightarrow b$ $\times$ $\overrightarrow r$ <br><br>$\Rightarrow$ $$\overrightarrow r \times \overrightarrow a = - \overrightarrow r \times \overrightarrow b $$<br><br>$\overrightarrow r \times (\overrightarrow a + \overrightarrow b ) = 0$<br><br>$\overrightarrow r ||(\overrightarrow a + \overrightarrow b )$<br><br>$\overrightarrow r = \lambda (\overrightarrow a + \overrightarrow b )$<br><br>$$(\overrightarrow a + \overrightarrow b = 3\widehat i - \widehat j + 2\widehat k)$$<br><br>$\because$ $\overrightarrow r \,.\,(2\widehat i + 5\widehat j - \alpha \widehat k) = - 1$<br><br>$$\lambda \left[ {3\widehat i - \widehat j + 2\widehat k} \right]\,.\,\left[ {2\widehat i + 5\widehat j - \alpha \widehat k} \right] = - 1$$<br><br>$\Rightarrow \lambda (6 - 5 - 2\alpha ) = - 1$<br><br>$\lambda (1 - 2\alpha ) = - 1$ .... (1)<br><br>$\overrightarrow r \,.\,(\alpha \widehat i + 2\widehat j + \widehat k) = 3$<br><br>$$\lambda (3\widehat i - \widehat j + 2\widehat k)\,.\,(\alpha \widehat i + 2\widehat j + \widehat k) = 3$$<br><br>$\Rightarrow \lambda [3\alpha - 2 + 2] = 3 \Rightarrow \lambda \alpha = 1$ .... (2)<br><br>From (1) &amp; (2)<br><br>$\lambda \left[ {1 - {2 \over \lambda }} \right] = - 1$<br><br>$\lambda - 2 = - 1 \Rightarrow \lambda = 1\,\alpha = 1$<br><br>$\overrightarrow r = 3\widehat i - \widehat j + 2\widehat k$<br><br>$\alpha + |\overrightarrow r {|^2} = 1 + 14 = 15$