We have to prove,

$\left|\begin{array}{ccc}y^2 z^2 & y z & y+z \\ z^2 x^2 & z x & z+x \\ x^2 y^2 & x y & x+y\end{array}\right|=0$

$$\begin{array}{r} \therefore\quad \mathrm{LHS}=\left|\begin{array}{rll} y^2 z^2 & y z & y+z \\ z^2 x^2 & z x & z+x \\ x^2 y^2 & x y & x+y \end{array}\right|=\frac{1}{x y z}\left|\begin{array}{rrr} x y^2 z^2 & x y z & x y+x z \\ x^2 y z^2 & x y z & y z+x y \\ x^2 y^2 z & x y z & x z+y z \end{array}\right| \\ {\left[\because R_1 \rightarrow x R_1, R_2 \rightarrow y R_2, R_3 \rightarrow z R_3\right]} \end{array}$$

$$=\frac{1}{x y z}(x y z)^2\left|\begin{array}{ccc} y z & 1 & x y+x z \\ x z & 1 & y z+x y \\ x y & 1 & x z+y z \end{array}\right|\quad$$ [taking $(x y z)$ common from $\mathrm{C}_1$ and $\mathrm{C}_2$]

$$\begin{aligned} & =x y z\left|\begin{array}{ccc} y z & 1 & x y+y z+z x \\ x z & 1 & x y+y z+z x \\ x y & 1 & x y+y z+z x \end{array}\right|\left[C_3 \rightarrow C_3+C_1\right] \\ & =x y z(x y+y z+z x)\left|\begin{array}{ccc} y z & 1 & 1 \\ x z & 1 & 1 \\ x y & 1 & 1 \end{array}\right| \end{aligned}$$

$$\begin{array}{lr} =0 & {\left[\text { taking }(x y+y z+z x) \text { common from } C_3\right]} \\ =\text { RHS } & {\left[\text { since, } C_2 \text { and } C_3 \text { are identicals }\right]} \\ \text { Hence proved. } \end{array}$$

We have to prove,

$\left|\begin{array}{ccc}y+z & z & y \\ z & z+x & x \\ y & x & x+y\end{array}\right|=4 x y z$

$$\begin{aligned} \mathrm{LHS} & =\left|\begin{array}{ccc} y+z & z & y \\ z & z+x & x \\ y & x & x+y \end{array}\right| \\ & =\left|\begin{array}{ccc} y+z+z+y & z & y \\ z+z+x+x & z+x & x \\ y+x+x+y & x & x+y \end{array}\right| \quad\left[\because C_1 \rightarrow C_1+C_2+C_3\right] \end{aligned}$$

$$=2\left|\begin{array}{ccc} (y+z) & z & y \\ (z+x) & z+x & x \\ (x+y) & x & x+y \end{array}\right|\quad$$ [taking 2 common from $\mathrm{C}_1$ ]

$$\begin{array}{ll} =2\left|\begin{array}{ccc} y & z & y \\ 0 & z+x & x \\ y & x & x+y \end{array}\right| & {\left[\because C_1 \rightarrow C_1-C_2\right]} \\ =2\left|\begin{array}{ccc} 0 & z-x & -x \\ 0 & z+x & x \\ y & x & x+y \end{array}\right| & {\left[\because R_1 \rightarrow R_1-R_3\right]} \end{array}$$

$$\begin{aligned} &\begin{aligned} & =2\left[y\left(x z-x^2+x z+x^2\right)\right] \\ & =4 x y z=\mathrm{RHS} \end{aligned}\\ &\text { Hence proved. } \end{aligned}$$

We have to prove,

$\left|\begin{array}{ccc}a^2+2 a & 2 a+1 & 1 \\ 2 a+1 & a+2 & 1 \\ 3 & 3 & 1\end{array}\right|=(a-1)^3$

$\therefore \quad$ LHS $=\left|\begin{array}{ccc}a^2+2 a & 2 a+1 & 1 \\ 2 a+1 & a+2 & 1 \\ 3 & 3 & 1\end{array}\right|$

$$\begin{aligned} =\left|\begin{array}{ccc} a^2+2 a-2 a-1 & 2 a+1-a-2 & 0 \\ 2 a+1-3 & a+2-3 & 0 \\ 3 & 3 & 1 \end{array}\right| & \\ & {\left[\because R_1 \rightarrow R_1-R_2 \text { and } R_2 \rightarrow R_2-R_3\right] } \end{aligned}$$

$$\begin{aligned} &=\left|\begin{array}{ccc} (a-1)(a+1) & (a-1) & 0 \\ 2(a-1) & (a-1) & 0 \\ 3 & 3 & 1 \end{array}\right|=(a-1)^2\left|\begin{array}{ccc} (a+1) & 1 & 0 \\ 2 & 1 & 0 \\ 3 & 3 & 1 \end{array}\right| \\ &\text { [taking } \left.(a-1) \text { common from } R_1 \text { and } R_2 \text { each] }\right] \end{aligned}$$

$$\begin{aligned} & =(a-1)^2[1(a+1)-2]=(a-1)^3 \\ & =\text { RHS }\quad\text{Hence proved.} \end{aligned}$$

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

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

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

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

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

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.