JEE Main 17 March 2021 Shift 2 Solved Paper

Given, slope of OA=tan‌α=‌

sin‌α

cos‌α

=‌

sin‌α−0

cos‌α−0

∴ Coordinates of A must be (cos‌α,sin‌α)
Similarly, B=(cos‌β,sin‌β) and C=(cos‌γ,sin‌γ)
Given, circumcentre coincides with origin and orthocentre lies on Y-axis.
So, centroid must lie on Y-axis because circumcentre, orthocentre and centroid of any triangle always lies on the same line.
∵cos‌α+cos‌β+cos‌γ=0 ...(i)
(because x-coordinate on Y-axis is zero)
Now, =4(cos3α+cos3β+cos3γ)−3×0 (from Eq. (i))
=12‌cos‌α⋅cos‌β⋅cos‌γ  (from Eq. (ii))