MCQ
Let $\left| {\begin{array}{*{20}{c}}
  {{{\left( {a - x} \right)}^2}}&{{{\left( {a - y} \right)}^2}}&{{{\left( {a - z} \right)}^2}} \\ 
  {{{\left( {b - x} \right)}^2}}&{{{\left( {b - y} \right)}^2}}&{{{\left( {b - z} \right)}^2}} \\ 
  {{{\left( {c - x} \right)}^2}}&{{{\left( {c - y} \right)}^2}}&{{{\left( {c - z} \right)}^2}} 
\end{array}} \right| = \frac{{ - 351}}{8}$ . If $x, y , z$ are the roots of the equation $8t^3 - 62t^2 + 43t -7 = 0$ and satisfy the determinant above, and $a, b, c$ are distinct number then value of $|(a - b) (b - c) (c - a)|$ is
  • $2$
  • B
    $4$
  • C
    $10$
  • D
    $14$

Answer

Correct option: A.
$2$
a
Roots of given cubic is $7, \frac{1}{2} \cdot \frac{1}{4}$

Given determinant is

$\left| {\begin{array}{*{20}{c}}
{{a^2}}&{ - 2a}&1\\
{{b^2}}&{2b}&1\\
{{c^2}}&{ - 2b}&1
\end{array}} \right| \times \left| {\begin{array}{*{20}{c}}
1&1&1\\
x&y&z\\
{{x^2}}&{{y^2}}&{{z^2}}
\end{array}} \right|$

$2(a-b)(b-c)(c-a)(x-y)(y-z)(z-x)$

Hence $|(a-b)(b-c)(c-a)|=2$

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