If Q(alpha, beta, gamma) is the harmonic conj | vedclass

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(A) Let the points be $A(2, -5, 3)$ and $B(-1, -8, 0)$. The point $P(0, -7, 1)$ divides the line segment $AB$ in the ratio $k:1$. Using the section fo

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$4$

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(A) Let the points be $A(2, -5, 3)$ and $B(-1, -8, 0)$. The point $P(0, -7, 1)$ divides the line segment $AB$ in the ratio $k:1$. Using the section formula: $0 = \frac{k(-1) + 1(2)}{k+1} \implies -k + 2 = 0 \implies k = 2$. So,$P$ divides $AB$ internally in the ratio $2:1$. Since $Q$ is the harmonic conjugate of $P$ with respect to $AB$,$Q$ divides $AB$ externally in the ratio $2:1$. Using the external section formula for $Q(\alpha, \beta, \gamma)$: $\alpha = \frac{2(-1) - 1(2)}{2-1} = -2 - 2 = -4$. $\beta = \frac{2(-8) - 1(-5)}{2-1} = -16 + 5 = -11$. $\gamma = \frac{2(0) - 1(3)}{2-1} = 0 - 3 = -3$. Thus,$Q = (-4, -11, -3)$. We need to find $\alpha - \beta + \gamma = -4 - (-11) + (-3) = -4 + 11 - 3 = 4$.

If $Q(\alpha, \beta, \gamma)$ is the harmonic conjugate of the point $P(0, -7, 1)$ with respect to the line segment joining the points $A(2, -5, 3)$ and $B(-1, -8, 0)$,then $\alpha - \beta + \gamma =$

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