Q102 / 160 · TS EAMCET 2022 (19 Jul Shift 2) · Mathematics

$\mathrm{A}(27,-243,81)$ is a point in space. $\mathrm{B}, \mathrm{C}, \mathrm{D}$ are images of A with respect to $\mathrm{XY}, \mathrm{YZ}$ and $\mathrm{ZX}$ planes respectively. If the centroid of the triangle $\mathrm{BCD}$ is $(\alpha, \beta, \gamma)$, then $\alpha+\beta+\gamma=$

  • A
    -25
  • B
    45
  • C
    25
  • Correct answer
    D
    -45

Solution

Given point $\mathrm{A}(27,-243,81)$ in space, now images of point $A$ in xyh plane, yzh plane and zx-planes are B $(27,-243,-81), C(-27,-243,81)$ and $D(27,243,81)$ respectively. Hence centroid of the triangle $\mathrm{BCD}$ is $=(9,-81,27) \equiv(\alpha, \beta, \gamma)$ hence, $\alpha+\beta+\gamma=-45$

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

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