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(B) Let $P(2, 3, 4)$ divide the line segment $AB$ in the ratio $k:1$. Using the section formula, the coordinates of $P$ are given by $\left(\frac{6k+3

Vedclass · Accepted Answer

$-\frac{3}{5}$

Vedclass · Answer

(B) Let $P(2, 3, 4)$ divide the line segment $AB$ in the ratio $k:1$. Using the section formula, the coordinates of $P$ are given by $\left(\frac{6k+3}{k+1}, \frac{-17k-2}{k+1}, \frac{-4k+2}{k+1}\right)$. Equating the $x$-coordinate: $\frac{6k+3}{k+1} = 2 \Rightarrow 6k+3 = 2k+2 \Rightarrow 4k = -1 \Rightarrow k = -\frac{1}{4}$. Since $P$ divides $AB$ internally in the ratio $k:1 = -\frac{1}{4}:1$, the harmonic conjugate $Q$ divides $AB$ externally in the ratio $k:1 = \frac{1}{4}:1$ (or internally in the ratio $\frac{1}{4}:1$). Using the section formula for $Q$ with ratio $\lambda = \frac{1}{4}$: $Q = \left(\frac{\frac{1}{4}(6) + 1(3)}{\frac{1}{4}+1}, \frac{\frac{1}{4}(-17) + 1(-2)}{\frac{1}{4}+1}, \frac{\frac{1}{4}(-4) + 1(2)}{\frac{1}{4}+1}\right) = \left(\frac{1.5+3}{1.25}, \frac{-4.25-2}{1.25}, \frac{-1+2}{1.25}\right) = \left(\frac{4.5}{1.25}, \frac{-6.25}{1.25}, \frac{1}{1.25}\right) = \left(\frac{18}{5}, -5, \frac{4}{5}\right)$. Thus, $\alpha = \frac{18}{5}$, $\beta = -5$, and $\gamma = \frac{4}{5}$. Therefore, $\alpha + \beta + \gamma = \frac{18}{5} - 5 + \frac{4}{5} = \frac{22}{5} - \frac{25}{5} = -\frac{3}{5}$.

If the harmonic conjugate of $P(2, 3, 4)$ with respect to the line segment joining the points $A(3, -2, 2)$ and $B(6, -17, -4)$ is $Q(\alpha, \beta, \gamma)$, then $\alpha + \beta + \gamma =$

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