The Cartesian coordinates (x, y) of a point on a curve are given by $x : y : 1 = t^{3} : t^{2} - 3 : t - 1$ where t is a parameter, then the points given by t = a, b, c are collinear, if

abc + 3(a + b + c) = ab + bc + ca

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3abc + 2(a + b + c) = ab + bc + ca

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abc + 2(a + b + c) = 3(ab + bc + ca)

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None of the above

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Solution

The correct option is A abc + 3(a + b + c) = ab + bc + ca
Given $x = \frac{t^{3}}{t - 1}, y = \frac{t^{2} - 3}{t - 1},$ where t = a,b,c
If the three points are collinear, then they will satisfy the same equation Ax + By + C = 0.
$\Rightarrow A (\frac{t^{3}}{t - 1}) + B (\frac{t^{2} - 3}{t - 1}) + C = 0$
$\Rightarrow A t^{3} + B t^{2} + C t - (3 B + C) = 0,$ which is cubic equation in t
$∴ a + b + c = - \frac{B}{A}$
$a b + b c + c a = + \frac{C}{A}$
$a b c = + (\frac{3 B + C}{A})$
$\Rightarrow a b c = + 3 (\frac{B}{A}) + (\frac{C}{A})$
$a b c = - 3 (a + b + c) + a b + b c + c a$
$∴ a b + b c + c a = a b c + 3 (a + b + c)$

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