$$\begin{aligned} &\text { We have to prove, }\left|\begin{array}{ccc} 1 & \cos C & \cos B \\ \cos C & 1 & \cos A \\ \cos B & \cos A & 1 \end{array}\right|=0\\ &\therefore \quad \mathrm{LHS}=\left|\begin{array}{ccc} 1 & \cos C & \cos B \\ \cos C & 1 & \cos A \\ \cos B & \cos A & 1 \end{array}\right| \end{aligned}$$

$$\begin{aligned} & =1\left(1-\cos ^2 A\right)-\cos C(\cos C-\cos A \cdot \cos B)+\cos B(\cos C \cdot \cos A-\cos B) \\ & =\sin ^2 A-\cos ^2 C+\cos A \cdot \cos B \cdot \cos C+\cos A \cdot \cos B \cdot \cos C-\cos ^2 B \\ & =\sin ^2 A-\cos ^2 B+2 \cos A \cdot \cos B \cdot \cos C-\cos ^2 C \\ & =-\cos (A+B) \cdot \cos (A-B)+2 \cos A \cdot \cos B \cdot \cos C-\cos ^2 C \\ & \quad\left[\because \cos ^2 B-\sin ^2 A=\cos (A+B) \cdot \cos (A-B)\right] \end{aligned}$$

$$\begin{aligned} & =-\cos (-C) \cdot \cos (A-B)+\cos C(2 \cos A \cdot \cos B-\cos C) \quad[\because \cos (-\theta)=\cos \theta] \\ & =-\cos C(\cos A \cdot \cos B+\sin A \cdot \sin B-2 \cos A \cdot \cos B+\cos C) \\ & =\cos C(\cos A \cdot \cos B-\sin A \cdot \sin B-\cos C) \\ & =\cos C[\cos (A+B)-\cos C] \\ & =\cos C(\cos C-\cos C)=0=\text { RHS } \quad \text { Hence proved. } \end{aligned}$$

$$\begin{aligned} &\text { Since, we know that area of a triangle with vertices }\left(x_1, y_1\right),\left(x_2, y_2\right) \text { and }\left(x_3, y_3\right) \text {, is given by }\\ &\begin{aligned} \Delta & =\frac{1}{2}\left|\begin{array}{lll} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{array}\right| \\ \Rightarrow\quad \Delta^2 & =\frac{1}{4}\left|\begin{array}{lll} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{array}\right|\quad\text{.... (i)} \end{aligned} \end{aligned}$$

$$ \begin{aligned} &\text { We know that, area of an equilateral triangle with side } a \text {, }\\ &\begin{aligned} \Delta =\frac{1}{2}\left(\frac{\sqrt{3}}{2}\right) a^2=\frac{\sqrt{3}}{4} a^2 \\ \Rightarrow \quad \Delta^2 =\frac{3}{16} a^4\quad\text{.... (ii)} \end{aligned} \end{aligned}$$

From Eqs. (i) and (ii), $\quad \frac{3}{16} a^4=\frac{1}{4}\left|\begin{array}{lll}x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1\end{array}\right|^2$

$\Rightarrow \quad\left|\begin{array}{lll}x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1\end{array}\right|^2=\frac{3}{4} a^4\quad$ Hence proved.

We have, $$ \left|\begin{array}{rrc} 1 & 1 & \sin 3 \theta \\ -4 & 3 & \cos 2 \theta \\ 7 & -7 & -2 \end{array}\right|=0$$

$\Rightarrow \quad\left|\begin{array}{rrc}0 & 1 & \sin 3 \theta \\ -7 & 3 & \cos 2 \theta \\ 14 & -7 & -2\end{array}\right|=0 \quad\left[\because C_1 \rightarrow C_1-C_2\right]$

$$ \Rightarrow \quad 7\left|\begin{array}{ccc} 0 & 1 & \sin 3 \theta \\ -1 & 3 & \cos 2 \theta \\ 2 & -7 & -2 \end{array}\right|=0\quad$$ [taking 7 common from $\mathrm{C}_1$ ]

$\Rightarrow \quad 7[0-1(2-2 \cos 2 \theta)+\sin 3 \theta(7-6)]=0 \quad\left[\right.$ expanding along $\left.R_1\right]$

$$\begin{array}{lr} \Rightarrow & 7[-2(1-\cos 2 \theta)+\sin 3 \theta]=0 \\ \Rightarrow & -14+14 \cos 2 \theta+7 \sin 3 \theta=0 \\ \Rightarrow & 14 \cos 2 \theta+7 \sin 3 \theta=14 \\ \Rightarrow & 14\left(1-2 \sin ^2 \theta\right)+7\left(3 \sin \theta-4 \sin ^3 \theta\right)=14 \\ \Rightarrow & -28 \sin ^2 \theta+14+21 \sin \theta-28 \sin ^3 \theta=14 \\ \Rightarrow & -28 \sin ^2 \theta-28 \sin ^3 \theta+21 \sin \theta=0 \\ \Rightarrow & 28 \sin ^3 \theta+28 \sin ^2 \theta-21 \sin \theta=0 \\ \Rightarrow & 4 \sin ^3 \theta+4 \sin ^2 \theta-3 \sin \theta=0 \\ \Rightarrow & \sin \theta\left(4 \sin ^2 \theta+4 \sin \theta-3\right)=0 \\ \Rightarrow & \text { Either } \sin \theta=0, \\ \Rightarrow & \theta=n \pi \text { or } 4 \sin ^2 \theta+4 \sin \theta-3=0 \end{array}$$

$$ \begin{aligned} & \therefore \quad \sin \theta=\frac{-4 \pm \sqrt{16+48}}{8}=\frac{-4 \pm \sqrt{64}}{8} \\ & =\frac{-4 \pm 8}{8}=\frac{4}{8}, \frac{-12}{8} \\ & \sin \theta=\frac{1}{2}, \frac{-3}{2} \\ & \text { If } \quad \sin \theta=\frac{1}{2}=\sin \frac{\pi}{6} \text {, then } \\ & \theta=n \pi+(-1)^n \frac{\pi}{6} \\ & \text { Hence, } \quad \sin \theta=\frac{-3}{2}\quad \text { [not possible because }-1 \leq \sin \theta \leq 1 \text { ] } \end{aligned}$$

Given, $$\left|\begin{array}{lll} 4-x & 4+x & 4+x \\ 4+x & 4-x & 4+x \\ 4+x & 4+x & 4-x \end{array}\right|=0$$

$\Rightarrow \quad\left|\begin{array}{ccc}12+x & 12+x & 12+x \\ 4+x & 4-x & 4+x \\ 4+x & 4+x & 4-x\end{array}\right|=0 \quad\left[\because R_1 \rightarrow R_1+R_2+R_3\right]$

$\Rightarrow \quad(12+x)\left|\begin{array}{ccc}1 & 1 & 1 \\ 4+x & 4-x & 4+x \\ 4+x & 4+x & 4-x\end{array}\right|=0 \quad$ [taking $(12+x)$ common from $R_1$ ]

$$ \begin{aligned} \Rightarrow \quad & (12+x)\left|\begin{array}{ccc} 0 & 0 & 1 \\ 0 & 8 & 4+x \\ 2 x & 8 & 4-x \end{array}\right| =0 \quad\left[\because C_1 \rightarrow C_1-C_3 \text { and } C_2 \rightarrow C_2+C_3\right] \\ \Rightarrow \quad & (12+x)[1 \cdot(-16 x)] =0 \\ \Rightarrow \quad & (12+x)(-16 x) =0 \\ \therefore \quad & x =-12,0 \end{aligned}$$

We know that, $$a_{r+1}=A R^{(r+1)-1}=A ^r $$

where $r=r$ th term of a GP, A = First term of a GP and $R=$ Common ratio of GP

$$\begin{aligned} &\text { We have, }\\ &\begin{aligned} &\left|\begin{array}{ccc} a_{r+1} & a_{r+5} & a_{r+9} \\ a_{r+7} & a_{r+11} & a_{r+15} \\ a_{r+11} & a_{r+17} & a_{r+21} \end{array}\right| \\ &=\left|\begin{array}{ccc} A R^r & A R^{r+4} & A R^{r+8} \\ A R^{r+6} & A R^{r+10} & A R^{r+14} \\ A R^{r+10} & A R^{r+16} & A R^{r+20} \end{array}\right| \\ &=A R^r \cdot A R^{r+6} \cdot A R^{r+10}\left|\begin{array}{ccc} 1 & A R^4 & A R^8 \\ 1 & A R^4 & A R^8 \\ 1 & A R^6 & A R^{10} \end{array}\right| \end{aligned} \end{aligned}$$

$\begin{array}{r}\text { [taking } A R^r, A R^{r+6} \text { and } A R^{r+10} \text { common from } R_1, R_2 \text { and } R_3, \text { respectively] } \\ =0 \text { [since, } R_1 \text { and } R_2 \text { are identicals] }\end{array}$