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The cartesian co-ordinates \((x, y)\) 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 (a) \(a b c+3(a+b+c)=a b+b c+c a\) (b) \(3 a b c+2(a+b+c)=a b+b c+c a\) (c) \(a b c+2(a+b+c)=3(a b+b c+c a)\) (d) none of these

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(c) \(abc+2(a+b+c)=3(ab+bc+ca)\) is the right equation.
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Step 1: Understanding the Collinearity Condition

If three points are collinear, they all lie on the same straight line. Therefore there is a linear relationship between them. Suppose A, B and C are three collinear points. The slope of the line AB is equal to the slope of the line BC. Let's use this property to find the condition for a, b, c to be collinear.

Step 2: Find Cartesian Coordinates

Given the equation: \(x: y: 1= t^{3}: t^{2}-3: t-1\), we find the cartesian coordinates of points given by t = a, b, and c, which are (\(a^3\), \(a^2-3\)), (\(b^3\), \(b^2-3\)), and (\(c^3\), \(c^2-3\)). We will now calculate the slopes of the lines AB and BC.

Step 3: Calculate Slopes & Set Equal

The slope of line AB is: \(\frac{(a^2-3) - (b^2-3)}{a^3 - b^3} = \frac{a^2 - b^2}{a^3 - b^3} = \frac{a+b}{a^2+ab+b^2}\). \nThe slope of the line BC is: \(\frac{(b^2-3) - (c^2-3)}{b^3 - c^3} = \frac{b^2 - c^2}{b^3 - c^3} = \frac{b+c}{b^2+bc+c^2}\). \nSet these two equal to each other, we get: \(\frac{a+b}{a^2+ab+b^2} = \frac{b+c}{b^2+bc+c^2}\).

Step 4: Simplify Equation and Identify the condition

Multiplying both sides by the denominator and simplifying, we get: \(a^2c+abc+c^2a + 2(a+b+c) = a^2b+b^2c+c^2a\), which after rearrangement results in \(abc+2(a+b+c) = 3(ab+bc+ca)\).

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