Among the statements:I: If begin{vmatrix} 1 & | vedclass

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(A) For statement $I$: Let $\cos \alpha = x, \cos \beta = y, \cos \gamma = z$.The given equation is $\begin{vmatrix} 1 & x & y \ x & 1 & z \ y & z &

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both are false

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(A) For statement $I$: Let $\cos \alpha = x, \cos \beta = y, \cos \gamma = z$.The given equation is $\begin{vmatrix} 1 & x & y \ x & 1 & z \ y & z & 1 \end{vmatrix} = \begin{vmatrix} 0 & x & y \ x & 0 & z \ y & z & 0 \end{vmatrix}$.Expanding the left determinant: $1(1-z^2) - x(x-yz) + y(xz-y) = 1 - z^2 - x^2 + xyz + xyz - y^2 = 1 - (x^2+y^2+z^2) + 2xyz$.Expanding the right determinant: $0(0-z^2) - x(0-yz) + y(xz-0) = xyz + xyz = 2xyz$.Equating both sides: $1 - (x^2+y^2+z^2) + 2xyz = 2xyz \implies x^2+y^2+z^2 = 1$.Thus, $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1 
eq \frac{3}{2}$. Statement $I$ is false.For statement $II$: Let $f(x) = \begin{vmatrix} x^{2}+x & x+1 & x-2 \ 2x^{2}+3x-1 & 3x & 3x-3 \ x^{2}+2x+3 & 2x-1 & 2x-1 \end{vmatrix} = px+q$.Setting $x=0$: $q = \begin{vmatrix} 0 & 1 & -2 \ -1 & 0 & -3 \ 3 & -1 & -1 \end{vmatrix} = 0(0-3) - 1(1+9) - 2(1-0) = -10 - 2 = -12$.Setting $x=1$: $p+q = \begin{vmatrix} 2 & 2 & -1 \ 4 & 3 & 3 \ 6 & 1 & 1 \end{vmatrix} = 2(3-3) - 2(4-18) - 1(4-18) = 0 + 28 + 14 = 42$.Since $q = -12$, $p - 12 = 42 \implies p = 54$.Checking $p^2 = 196q^2$: $54^2 = 2916$ and $196(-12)^2 = 196 	imes 144 = 28224$.Since $2916 
eq 28224$, statement $II$ is false.

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