Mathematics - Vector Question with Solution | TestHub

$\left|\begin{array}{ccc}a+1& a& a\\ a& a+1& a\\ a& a& a+1\end{array}\right|=0\Rightarrow a+1+a+a=0\Rightarrow a=-\frac{1}{3}$
$\overrightarrow{P}=\frac{2}{3}\widehat{i}-\frac{1}{3}\widehat{j}-\widehat{k}$
$\overrightarrow{Q}=\frac{1}{3}\left(-\widehat{i}+2\widehat{j}-\widehat{k}\right)$
$\overrightarrow{R}=\frac{1}{3}\left(-\widehat{i}-\widehat{j}+2\widehat{k}\right)$
$\overrightarrow{P}.\overrightarrow{Q}=\frac{1}{9}\left(-2-2+1\right)=-\frac{1}{3}$
$\overrightarrow{R}\times \overrightarrow{Q}=\frac{1}{9}\left|\begin{array}{ccc}i& j& k\\ -1& 2& -1\\ -1& -1& 2\end{array}\right|=\frac{1}{9}\left(i\left(4-1\right)-j\left(-2-1\right)+k\left(1+2\right)\right)=\frac{1}{9}\left(3i+3j+3k\right)=\frac{i+j+k}{3}$
$\left|\overrightarrow{R}\times \overrightarrow{Q}\right|=\frac{1}{3}\sqrt{3}\Rightarrow {\left|\overrightarrow{R}\times \overrightarrow{Q}\right|}^2=\frac{1}{3}$$3{\left(\overrightarrow{P}.\overrightarrow{Q}\right)}^2-\lambda {\left|\overrightarrow{R}\times \overrightarrow{Q}\right|}^2=0$
$3.\frac{1}{9}-\lambda .\frac{1}{3}=0\Rightarrow \lambda =1$