Among the statements : I: If | … |, then cos ^2 α+cos ^2 β+cos ^2 γ= 3 2, and II : If | … |= x+ q, then ^2=196 q^2,

Statement I: Let $D_1 = \begin{vmatrix} 1 & \cos\alpha & \cos\beta \\ \cos\alpha & 1 & \cos\gamma \\ \cos\beta & \cos\gamma & 1 \end{vmatrix} = 1 - \cos^2\alpha - \cos^2\beta - \cos^2\gamma + 2\cos\alpha\cos\beta\cos\gamma$D1​=​1cosαcosβ​cosα1cosγ​cosβcosγ1​​=1−cos2α−cos2β−cos2γ+2cosαcosβcosγ.

$D_2 = \begin{vmatrix} 0 & \cos\alpha & \cos\beta \\ \cos\alpha & 0 & \cos\gamma \\ \cos\beta & \cos\gamma & 0 \end{vmatrix} = 2\cos\alpha\cos\beta\cos\gamma$D2​=​0cosαcosβ​cosα0cosγ​cosβcosγ0​​=2cosαcosβcosγ.

$D_1 = D_2 \Rightarrow 1 - \cos^2\alpha - \cos^2\beta - \cos^2\gamma = 0 \Rightarrow \cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1 \neq \frac{3}{2}$D1​=D2​⇒1−cos2α−cos2β−cos2γ=0⇒cos2α+cos2β+cos2γ=1=23​. Statement I is false.

Statement II: Evaluate by substituting $x = 0$x=0: determinant $= -12$=−12, so $q = -12$q=−12.

At $x = 1$x=1: determinant $= 12$=12, so $p + q = 12 \Rightarrow p = 24$p+q=12⇒p=24.

$p^2 = 576$p2=576 and $196q^2 = 196 \times 144 = 28224$196q2=196×144=28224. Since $p^2 \neq 196q^2$p2=196q2, Statement II is false.

Both statements are false.