Let the position vectors of three vertices of a triangle be $4 \vec{p} + \vec{q} - 3 \vec{r}, - 5 \vec{p} + \vec{q} + 2 \vec{r}$ and $2 \vec{p} - \vec{q} + 2 \vec{r}$ . If the position vectors of the orthocenter and the circumcenter of the triangle are $\frac{\vec{p} + \vec{q} + \vec{r}}{4} and α \vec{p} + β \vec{q} + γ \vec{r}$ respectively, then $α + 2 β + 5 γ$ is equal to :

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Detailed Solution

We know that

$\begin{array}{l} O (orthocentre) \frac{\vec{p} + \vec{q} + \vec{r}}{4} \\ C (circum centre) α \vec{p} + β \vec{q} + γ \vec{r} \\ C (centroid) = \frac{\vec{p} + \vec{q} + \vec{r}}{3} \end{array}$

by relation
$\begin{array}{l} \Rightarrow 2 (α \vec{p} + β \vec{q} + γ \vec{r}) + \frac{\vec{p} + \vec{q} + \vec{r}}{4} = 3 (\frac{\vec{p} + \vec{q} + \vec{r}}{3}) \\ \Rightarrow 8 (α \vec{p} + β \vec{q} + γ \vec{r}) = 3 (\vec{p} + \vec{q} + \vec{r}) \\ \Rightarrow 8 α = 3, 8 β = 3, 8 γ = 3 \\ α = \frac{3}{8}, β = \frac{3}{8}, γ = \frac{3}{8} \\ ∴ α + 2 β + 3 γ \\ \frac{3}{8} + \frac{6}{8} + \frac{15}{8} = \frac{24}{8} = 3 \end{array}$