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Text Solution
Solution : Let the required point be `(x_(1),y_(1))`, <br> Now, `3y=6-5x^(3)` or `(dy)/(dx):|_((x_(1),y_(1)))=2-5x_(1)^(2)` <br> the equation of the normal at `(x_(1),y_(1))` is <br> `y-y_(1)=(-1)/(2-5x_(1)^(2))(x-x_(1))` <br> If it passes through the origin, then <br> `0-y_(1)=(-1)/(2-5x_(1)^(2))(0-x_(1))` <br> Since `(x_(1),y_(1))` lies on the given curve, <br> `3y_(1)=6x_(1)-5x_(1)^(3)` <br> Solving equation (1) and (2) we obtain `x_(1)=1` and `y_(1)=1/3` <br> Hence, the required point is `(1,1/3)` so, `p=1, q=1, r=3`