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Q. Let medians $L_1, L_2$ and $L_3$ of a $\triangle A B C$ belong to the family of lines $2 x-y+2+\lambda(x+2 y+$ $1)=0$ where $\lambda$ is a parameter. Points $P, Q, R$ are the mid-points of sides $B C, C A$ and $A B$ respectively where $P \equiv(2,6)$ and $Q=(-5,2)$. If area of $\triangle B R G$ is $6 \lambda$ where $G$ is centroid of $\triangle A B C$, then find the value of $\lambda$.

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Solution:

$2 x - y +2+\lambda( x +2 y +1)=0$
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$\operatorname{Ar}(\Delta BRG )=\operatorname{Ar}(\Delta AQG )=\frac{1}{2}\begin{vmatrix}-7 & -12 & 1 \\ -1 & 0 & 1 \\ -5 & 2 & 1\end{vmatrix}$
$\Rightarrow 30=6 \lambda \Rightarrow \lambda=5$