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Solving equations of the lines two at a time we get the vertices of the given triangle as A (- 7, 5), B (\frac{1}{3}, \frac{1}{9}) and C (\frac{5}{4}, \frac{7}{8}) Let P (α, α^{2}) be a point inside the triangle ABC . Since A and P lie on the same side of the line 5 x - 6 y - 1 = 0, both 5 - (7) - 6 (5) - 1 and 5 α - 6 α^{2} - 1 must have same sign . \Rightarrow 5 α - 6 α^{2} - 1 < 0 or 6 α - 5 α + 1 > 0 \Rightarrow (3 α - 1) (2 α - 1) > 0 \Rightarrow Either < \frac{1}{3} or α > \frac{1}{2} . . . . (1) Again since B and P lie on the same side of the line x + 2 y - 3 = 0, (\frac{1}{3}) + (\frac{2}{9}) and α + 2 α^{2} - 3 have the same sgn . 2 α^{2} + α - 3 < 0 \Rightarrow (2 α + 3) (α - 1) < 0 \Rightarrow \frac{- 3}{2} < α < 1 . . . (2) Lastly since C and P lie on the same side of the line 2 x + 3 y - 1 = 0 . 2 \times (\frac{5}{4}) + 3 \times (\frac{7}{8}) - 1 and 2 α + 3 α^{2} - 1 have the same sign . \Rightarrow 3 α^{2} + 2 α - 1 > 0 \Rightarrow (3 α - 1) (α + 1) > 0 \Rightarrow α < - 1 or > \frac{1}{3} . . . (3) Now, (1), (2) and (3) hold simultaneously i \frac{- 3}{2} < α < - 1 or \frac{1}{2} < α < 1 .

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