Complete the squares:

\((x^2 - 4x) - 3(y^2 + 2y) = 11\)
\((x-2)^2 - 4 - 3(y+1)^2 + 3 = 11\)
\((x-2)^2 - 3(y+1)^2 = 12 \Rightarrow \frac{(x-2)^2}{12} - \frac{(y+1)^2}{4} = 1\)

This is a hyperbola with \(a^2 = 12\), \(b^2 = 4\), centre \((2, -1)\).

• (A) Latus rectum = \(\frac{2b^2}{a} = \frac{8}{2\sqrt{3}} = \frac{4}{\sqrt{3}}\). True.
• (B) \(e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{1}{3}} = \frac{2}{\sqrt{3}}\). True.
• (C) The asymptotes are perpendicular only when \(a = b\) (rectangular hyperbola). Here \(a \ne b\). False.
• (D) The centre is \((2, -1)\), not \((1, -2)\). False.

So both C and D are false.

Answer: (E) Both C and D are wrong