Since a, b, c are in A.P.,
b-a = c-b = d, common difference....(1)
Now \(\Delta=\begin{vmatrix} x+1&x+2&x+a\\x+2&x+3&x+b\\x+3&x+4&x+c\end{vmatrix}\)
= \(\begin{vmatrix}x+1&x+2&x+a\\1&1&b-a\\1&1&c-b \end{vmatrix}\)
= \(\begin{vmatrix}x+1&x+2&x+a\\1&1&d\\1&1&d \end{vmatrix}\)
= 0 Which is true

If A be the first term and R, the common ratio of G.P, then a = AR^p-1, b = AR^q-1 and c = AR^r-1
Taking logs,
\(\left \{\begin{matrix} loga=logA+(p-1)logR, \\ logb=logA+(q-1)logR \\ logc=logA+(r-1)logR \end{matrix} \right\} \ \ .......(1)\)
= \(\begin{vmatrix} loga&p&1\\logb&q&1\\logc&r&1\end{vmatrix}\)
= \(\begin{vmatrix}logA+(p-1)logR& p&1\\logA+(q-1)logR&q&1\\ logA+(r-1)logR&r&1\end{vmatrix}\)
= \(\begin{vmatrix} loga&p&1\\logb&q&1\\logc&r&1\end{vmatrix}\)+\(\begin{vmatrix} (p-1)&p&1\\(q-1)&q&1\\(r-1)&r&1\end{vmatrix}\)
= \(logA\begin{vmatrix}1&p&1\\1&q&1\\1&r&1 \end{vmatrix}+logR\begin{vmatrix} p-1&p&1\\q-1&q&1\\r-1&r&1\end{vmatrix}\)
= \(logA(0)+logR\begin{vmatrix} 0&p&1\\0&q&1\\0&r&1\end{vmatrix}\)
= 0 + 0 which is true.

\(\Delta=\left| \begin{matrix} { a }^{ 2 } &-{ (b-c) }^{ 2 } & bc \\ { b }^{ 2 } &-{ (c-a) }^{ 2 } & ca \\ { c }^{ 2 } &-{ (a-b) }^{ 2 } & ab \end{matrix} \right| \)
= \(-\left| \begin{matrix} { a }^{ 2 } &{ (b-c) }^{ 2 } & bc \\ { b }^{ 2 } &{ (c-a) }^{ 2 } & ca \\ { c }^{ 2 } &{ (a-b) }^{ 2 } & ab \end{matrix} \right| \)
= \(\begin{vmatrix} a^2&a^2+b^2+c^2&bc\\b^2&a^2+b^2+c^2&ca\\c^2&a^2+b^2+c^2&ab\end{vmatrix}\)
= \(-(a^2+b^2+c^2)\begin{vmatrix} a^2&1&bc\\b^2&1&ca\\c^2&1&ab\end{vmatrix}\)
= \(-(a^2+b^2+c^2)\begin{vmatrix}a^2-b^2&0&c(b-a)\\b^2-c^2&0&a(c-b)\\c^2&1&ab \end{vmatrix}\)
= \((a^2+b^2+c^2(a-b)(b-c)\begin{vmatrix}a+b&0&c\\b+c&0& -a\\c^2&1&ab \end{vmatrix}\)
\(=-(a^2+b^2+c^2)(a-b)(b-c)(-1)(a^2-ab+bc+c^2)\)
= \((a-b)(b-c)(a^2+b^2+c^2)(c-a)(c+a+b)\)
= \((b-c)(c-a)(a+b+c)({ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 })\)

\(\Delta=\left| \begin{matrix} { yz }-x^{ 2 } & { zx }-y^{ 2 } & { xy-z }^{ 2 } \\ { zx-y }^{ 2 } & { xy }-z^{ 2 } & { yz-x }^{ 2 } \\ { xy-z }^{ 2 } & { yz }-x^{ 2 } & { zx-x }^{ 2 } \end{matrix} \right| \)
\(=\begin{vmatrix} -(x^2+y^2+z^2-xy-yz-zx)&zx-y^2&xy-z^2\\-(x^2+y^2+z^2-xy-yz-zx)&xy-z^2&yz-x^2\\-(x^2+y^2+z^2-xy-yz-zx)&yz-x^2&zx-y^2\end{vmatrix}\) [Operating C₁-->C₁+C₂+C₃]
\(= -(x^2+y^2+z^2-xy-yz-zx)\begin{vmatrix}1&zx-y^2&xy-z^2\\1&xy-z^2&yz-x^2\\1&yz-x^2&zx-y^2\end{vmatrix}\)  [Taking -(x²+y²+z²-xy-yz-zx) common from C₁]
=-(x²+y²+z²-xy-yz-zx)
\(\begin{vmatrix}0&(x-y)(x+y+z)&(y-z)(x+y+z)\\0&(x-y)(x+y+z )&(y-z)(x+y+z)\\1&yz-x^2&zx-y^2 \end{vmatrix}\) [Operating R₁-->R₁-R₃ and R₂+R₂-R₃]
\(=-(x^2+y^2+z^2-xy-yz-zx)(x+y +z)\)
\(=\begin{vmatrix} 0&x-y&y-z\\0&x-z&y-x\\1&yz-x^2&zx-y^2\end{vmatrix}\) [Taking (x+y+z)common from C₂and C₃]
\(=-(x+y+z)(x^3+y^3+z^3-3xyz)\begin{vmatrix}0&x-y&y-z\\0&x-z&y-x \\1&yz-x^2&zx-y^2\end{vmatrix}\)
\(=-(x+y+z)(x^3+y^3+z^3-3xyz)[(z-y)(y-x)-(x-z)(x-2)]\)
\(=-(x+y+z)(x^3+y^3+z^3-3xyz)[(yz-zx-y^2+xy-x^2-zx+zx-z^2)]\) [Expanding by C₁]
 \(=-(x+y+z)(x^3+y^3+z^3-3xyz)[(x^2+y^2+z^2-xy-yz-zx)]\)
Hence, △ is divisible by (x+y+z) and quotient is (x³+y³+z³-3xyz)(x²+y²+z²-x-yz-zx)

\(A={1\over2}\begin{vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1 \end{vmatrix}\)
But in equilateral triangle of each side equal toa, area = \(\sqrt{3}a^2\over4\)
= \({\sqrt{3}a^2\over4}={1\over2}\begin{vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1 \end{vmatrix}\)
\(\Rightarrow\begin{vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1 \end{vmatrix}={\sqrt{3}a^2\over2}\)
Hence,\(\left| \begin{matrix} x_1 &y_1 &1 \\x_2 &y_2 &1 \\x_3 &y_3 &1 \end{matrix} \right| ^2={3\over4}a^4\) is true.