Question

Let $cos\alpha = x$

$$\begin{matrix} & cos\beta = y \\ & cos\gamma = z \end{matrix}$$

Expending both sides, we get

$$x^{2} + y^{2} + z^{2} = 1$$

i.e. $\cos^{2}\alpha + \cos^{2}\beta + \cos^{2}\gamma = 1$

Statement 1 is false

Now,

$$\left| \begin{matrix} x^{2} + x & 1 + x & x - 2 \\ 2x^{2} + 3x - 1 & 3x & 3x - 3 \\ x^{2} + 2x + 3 & 2x - 1 & 2x - 1 \end{matrix} \right| = px + q $$Put $x = 0$ both sides

$${q = \left| \begin{matrix} 0 & 1 & - 2 \\ - 1 & 0 & - 3 \\ 3 & - 1 & - 1 \end{matrix} \right| }{\Rightarrow q = - 12 }$$Now put $x = 1$ both sides

$${p + q = \left| \begin{matrix} 2 & 2 & - 1 \\ 4 & 3 & 3 \\ 6 & 1 & 1 \end{matrix} \right| = 42 }{\Rightarrow p = 54 }$$Now $\frac{p^{2}}{q^{2}} = \left( \frac{54}{- 12} \right)^{2} + 196$

$$\Rightarrow p^{2} \neq 196q^{2} $$Statement (2) is false

Correct option (1)