Among the statements :
I: If $$ \begin{vmatrix}1 & \cos\alpha & \cos\beta \\\mathbf{\cos\alpha} & 1 & \mathbf{\cos\gamma} \\\mathbf{\cos\beta} & \mathbf{\cos\gamma} & 1\end{vmatrix}=\begin{vmatrix}0 & \mathbf{\cos\alpha}&\mathbf{\cos\beta} \\\mathbf{\cos\alpha} & 0 & \mathbf{\cos\gamma} \\\mathbf{\cos\beta} & \mathbf{\cos\gamma} & 0\end{vmatrix}$$, then $$\cos^{2}\alpha+\cos^{2}\beta+\cos^{2}\gamma=\frac{3}{2}$$, and 

II: $$\begin{vmatrix}x^{2}+x & x+1 & x-2 \\2x^{2}+3x-1 & 3x & 3x-3 \\x^{2}+2x+3 & 2x-1 & 2x-1\end{vmatrix} = px + q$$, then $$p^{2}=196q^{2}$$

Statement I

Put $$a=\cos\alpha,\;b=\cos\beta,\;c=\cos\gamma$$.
The first determinant becomes

$$\Delta_1=\begin{vmatrix}1&a&b\\a&1&c\\b&c&1\end{vmatrix}$$

For any symmetric matrix of this type the standard expansion gives
$$\Delta_1=1+2abc-a^{2}-b^{2}-c^{2}\;.\;-(1)$$

The second determinant is

$$\Delta_2=\begin{vmatrix}0&a&b\\a&0&c\\b&c&0\end{vmatrix}$$

Expanding along the first row:

$$\Delta_2=0\cdot\Bigl|\begin{smallmatrix}0&c\\c&0\end{smallmatrix}\Bigr| -a\Bigl|\begin{smallmatrix}a&c\\b&0\end{smallmatrix}\Bigr| +b\Bigl|\begin{smallmatrix}a&0\\b&c\end{smallmatrix}\Bigr| = -a(-bc)+b(ac)=2abc\;.\;-(2)$$

Given $$\Delta_1=\Delta_2$$, substitute $$(1)$$ and $$(2)$$:

$$1+2abc-a^{2}-b^{2}-c^{2}=2abc \;\;\Longrightarrow\;\;1-a^{2}-b^{2}-c^{2}=0$$

Hence $$\cos^{2}\alpha+\cos^{2}\beta+\cos^{2}\gamma=a^{2}+b^{2}+c^{2}=1$$, not $$\frac{3}{2}$$.
So Statement I is false.

Statement II

Let

$$D(x)=\begin{vmatrix}x^{2}+x & x+1 & x-2\\ 2x^{2}+3x-1 & 3x & 3x-3\\ x^{2}+2x+3 & 2x-1 & 2x-1\end{vmatrix}$$

Use the cofactor rule $$D=A_1E_1-B_1E_2+C_1E_3$$ with
$$(A_1,B_1,C_1)=(x^{2}+x,\;x+1,\;x-2).$$

$$E_1=\begin{vmatrix}0&3x-3\\2x-1&0\end{vmatrix}=3(2x-1)=6x-3$$
$$\;\;\Longrightarrow\;A_1E_1=(x^{2}+x)(6x-3)=6x^{3}+3x^{2}-3x$$

$$E_2=\begin{vmatrix}2x^{2}+3x-1&3x-3\\x^{2}+2x+3&2x-1\end{vmatrix} =x^{3}+x^{2}-8x+10$$
$$\;\;\Longrightarrow\;B_1E_2=(x+1)(x^{3}+x^{2}-8x+10) =x^{4}+2x^{3}-7x^{2}+2x+10$$

$$E_3=\begin{vmatrix}2x^{2}+3x-1&3x\\x^{2}+2x+3&2x-1\end{vmatrix} =x^{3}-2x^{2}-14x+1$$
$$\;\;\Longrightarrow\;C_1E_3=(x-2)(x^{3}-2x^{2}-14x+1) =x^{4}-4x^{3}-10x^{2}+29x-2$$

Therefore

$$D(x)=\bigl(6x^{3}+3x^{2}-3x\bigr) -\bigl(x^{4}+2x^{3}-7x^{2}+2x+10\bigr) +\bigl(x^{4}-4x^{3}-10x^{2}+29x-2\bigr)$$

Simplifying term by term:

$$D(x)=24x-12=12(2x-1)$$

Thus $$p=24,\;q=-12$$ and
$$p^{2}=24^{2}=576,\quad 196q^{2}=196\cdot144=28224\neq576.$$

Hence Statement II is also false.

Both statements are false  →  Option B.