Let the position vectors of three vertices of a \triangle be 4vecp+vecq-3vecr$4\vec{p}+\vec{q}-3\vec{r}$, -5vecp+vecq+2vecr$-5\vec{p}+\vec{q}+2\vec{r}$ and 2vecp-vecq+2vecr$2\vec{p}-\vec{q}+2\vec{r}$ If the position vectors of the orthocenter and the circumcenter of the \triangle are fracvecp+vecq+vecr4$\frac{\vec{p}+\vec{q}+\vec{r}}{4}$ and alphavecp+betavecq+gammavecr$\alpha\vec{p}+\beta\vec{q}+\gamma\vec{r}$ respectively, then alpha+2beta+5gamma$\alpha+2\beta+5\gamma$ is equal to: [cite: 3266, 3267, 3268, 3269, 3270, 3271, 3272]

### Related Formula 1. Centroid (G$G$) of a \triangle with vertices A, B, C$A, B, C$ is given by: vecG = fracvecA + vecB + vecC3$$\vec{G} = \frac{\vec{A} + \vec{B} + \vec{C}}{3}$$ 2. Euler\'s line property: The orthocenter (O$O$), centroid (G$G$), and circumcenter (C$C$) are collinear, and G$G$ divides the segment OC$OC$ internally in the ratio 2:1$2:1$. ### Step 1: Compute the Centroid Vector Sum the vectors of the three given vertices [cite: 3266, 3268]: vecA = 4vecp+vecq-3vecr$$\vec{A} = 4\vec{p}+\vec{q}-3\vec{r}$$ vecB = -5vecp+vecq+2vecr$$\vec{B} = -5\vec{p}+\vec{q}+2\vec{r}$$ vecC = 2vecp-vecq+2vecr$$\vec{C} = 2\vec{p}-\vec{q}+2\vec{r}$$ vecG = frac(4 - 5 + 2)vecp + (1 + 1 - 1)vecq + (-3 + 2 + 2)vecr3 = fracvecp + vecq + vecr3$$\vec{G} = \frac{(4 - 5 + 2)\vec{p} + (1 + 1 - 1)\vec{q} + (-3 + 2 + 2)\vec{r}}{3} = \frac{\vec{p} + \vec{q} + \vec{r}}{3}$$ ### Step 2: Apply Euler Line Section Ratio Using the section formula ratio O-G-C$O-G-C$ as 2:1$2:1$ [cite: 3931, 3932]:

vecG = frac2vecC + vecO3 Rightarrow 3vecG = 2vecC + vecO$$\vec{G} = \frac{2\vec{C} + \vec{O}}{3} \Rightarrow 3\vec{G} = 2\vec{C} + \vec{O}$$ 2vecC = 3vecG - vecO = 3left(fracvecp + vecq + vecr3right) - fracvecp + vecq + vecr4$$2\vec{C} = 3\vec{G} - \vec{O} = 3\left(\frac{\vec{p} + \vec{q} + \vec{r}}{3}\right) - \frac{\vec{p} + \vec{q} + \vec{r}}{4}$$ 2vecC = (vecp + vecq + vecr) - frac14(vecp + vecq + vecr) = frac34(vecp + vecq + vecr)$$2\vec{C} = (\vec{p} + \vec{q} + \vec{r}) - \frac{1}{4}(\vec{p} + \vec{q} + \vec{r}) = \frac{3}{4}(\vec{p} + \vec{q} + \vec{r})$$ vecC = frac38vecp + frac38vecq + frac38vecr$$\vec{C} = \frac{3}{8}\vec{p} + \frac{3}{8}\vec{q} + \frac{3}{8}\vec{r}$$ ### Step 3: Coefficient Matching Compare with the given circumcenter format alphavecp + betavecq + gammavecr$\alpha\vec{p} + \beta\vec{q} + \gamma\vec{r}$ [cite: 3270, 3939]: alpha = frac38, quad beta = frac38, quad gamma = frac38$$\alpha = \frac{3}{8}, \quad \beta = \frac{3}{8}, \quad \gamma = \frac{3}{8}$$ Calculate alpha + 2beta + 5gamma$\alpha + 2\beta + 5\gamma$ [cite: 3272, 3949]: frac38 + 2left(frac38right) + 5left(frac38right) = frac3 + 6 + 158 = frac248 = 3$$\frac{3}{8} + 2\left(\frac{3}{8}\right) + 5\left(\frac{3}{8}\right) = \frac{3 + 6 + 15}{8} = \frac{24}{8} = 3$$ ### Pattern Recognition Euler line configuration is universally O-G-C$O-G-C$ in 2:1$2:1$. Remember the mnemonic 'Oil-Gas-Company' or simply 3G = 2C + O$3G = 2C + O$ to prevent swapping structural coefficients under exam stress. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Vector Algebra Class 11 Mathematics: Properties of Triangles