If $C= 2 cos θ,$ then the value of the determinant

$Δ=\begin{vmatrix}C & 1 & 0\\1 & C & 1\\6 & 1 & C\end{vmatrix}$, is

none of these

The correct answer is option (4) : none of these

We have,

$Δ=\begin{vmatrix}C & 1 & 0\\1 & C & 1\\6 & 1 & C\end{vmatrix}$

$⇒Δ=\begin{vmatrix}0 & 1 & 0\\1-C^2 & C & 1\\6-C & 1 & C\end{vmatrix}$    [Applying $C_1→C_1-CC_2$]

$⇒Δ=\begin{vmatrix}1-C^2 & 1\\6-C& C\end{vmatrix} $     [Expanding along $R_1$]

$⇒Δ = -(C-C^3-6+C)$

$⇒Δ=C^3-2C+6$

$⇒Δ=8cos^3\theta - 4 cos \theta + 6 $

Hence, option (4) is correct.