Let A(3, 2, -4) and B(9, 8, -10) be two point | vedclass

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(B) The coordinates of $P_1$ dividing $AB$ in the ratio $1:2$ are given by the section formula:$P_1 = \left( \frac{1(9) + 2(3)}{1+2}, \frac{1(8) + 2(2

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$2$

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(B) The coordinates of $P_1$ dividing $AB$ in the ratio $1:2$ are given by the section formula:$P_1 = \left( \frac{1(9) + 2(3)}{1+2}, \frac{1(8) + 2(2)}{1+2}, \frac{1(-10) + 2(-4)}{1+2} \right) = \left( \frac{15}{3}, \frac{12}{3}, \frac{-18}{3} \right) = (5, 4, -6)$.The coordinates of $P_2$ dividing $AB$ in the ratio $2:1$ are:$P_2 = \left( \frac{2(9) + 1(3)}{2+1}, \frac{2(8) + 1(2)}{2+1}, \frac{2(-10) + 1(-4)}{2+1} \right) = \left( \frac{21}{3}, \frac{18}{3}, \frac{-24}{3} \right) = (7, 6, -8)$.Since $P(\alpha, \beta, \gamma)$ is the midpoint of $P_1 P_2$ (ratio $1:1$):$P = \left( \frac{5+7}{2}, \frac{4+6}{2}, \frac{-6-8}{2} \right) = (6, 5, -7)$.Thus, $\alpha = 6$, $\beta = 5$, and $\gamma = -7$.Calculating the expression:$\alpha + 2\beta + 2\gamma = 6 + 2(5) + 2(-7) = 6 + 10 - 14 = 2$.

Let $A(3, 2, -4)$ and $B(9, 8, -10)$ be two points. Let $P_1$ divide $AB$ in the ratio $1:2$ and $P_2$ divide $AB$ in the ratio $2:1$. If the point $P(\alpha, \beta, \gamma)$ divides $P_1 P_2$ in the ratio $1:1$, then $\alpha + 2\beta + 2\gamma =$

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