Discussion

Explanation:

P be the point of intersection of the lines 3x + 4y = 2a and 7x + 2y = 2018.

Solving these two equations we get the coordinates of P in terms of a.

∴ P ≡ $\left(\frac{4036-2a}{11},\frac{7a-3027}{11}\right)$

Q the point of intersection of the lines 3x + 4y = 2018 and 5x + 3y = 1

Solving these two equations we get the coordinates of Q.

∴ Q ≡ (-550, 917)

Now slope of line connecting P and Q is 2

∴ $\frac{{\displaystyle \frac{7a-3027}{11}}-917}{{\displaystyle \frac{4036-2a}{11}}+550}$ = $\frac{2}{1}$

⇒ $\frac{7a-13114}{10086-2a}$ = $\frac{2}{1}$

⇒ 11a = 33286

⇒ a = 3026

Hence, option (e).

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