Q98 / 160 · TS EAMCET 2024 (10 May Shift 2) · Mathematics

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

  • A
    $-\frac{2}{5}$
  • Correct answer
    B
    $-\frac{3}{5}$
  • C
    $\frac{7}{5}$
  • D
    $\frac{8}{5}$

Solution

Let $P(2,3,4)$ divides $A(3,-2,2) 7$ B $(6,-17,-4)$ in the ratio $\mathrm{k}: 1$ then $2=\frac{6 k+3}{k+1} \Rightarrow k=-\frac{1}{4}$
Harmonic conjugate Q divides in the ratio $k: 1$ i.e; $\frac{1}{4}: 1$ $Q(\alpha, \beta, \gamma)=\left(\frac{18}{5},-5, \frac{4}{5}\right)$
So, $\alpha+\beta+\gamma=\frac{-3}{5}$

Hence, the correct answer is option: 'B'.

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