Explanation:
(D) : Given,
\(\left|\begin{array}{ccc} a^{2}+1 & a b & a c \\ a b & b^{2}+1 & b c \\ a c & c b & c^{2}+1 \end{array}\right|\)
\(=\frac{1}{a b c}\left|\begin{array}{ccc} a\left(a^{2}+1\right) & a^{2} b & a^{2} c \\ a b^{2} & b\left(b^{2}+1\right) & b^{2} c \\ c^{2} a & c^{2} b & c\left(c^{2}+1\right) \end{array}\right|\)
On applying,
\(\left(\mathrm{R}_{1} \rightarrow \mathrm{aR} \mathrm{R}_{1}, \mathrm{R}_{2} \rightarrow \mathrm{bR}_{2}, \mathrm{R}_{3} \rightarrow \mathrm{cR}_{3}\right)\)
\(=\left|\begin{array}{ccc} \mathrm{a}^{2}+1 & \mathrm{a}^{2} & \mathrm{a}^{2} \\ \mathrm{~b}^{2} & \mathrm{~b}^{2}+1 & \mathrm{~b}^{2} \\ \mathrm{c}^{2} & \mathrm{c}^{2} & \mathrm{c}^{2}+1 \end{array}\right|\)
On applying,
\(\left(\mathrm{c}_{1} \rightarrow \frac{1}{\mathrm{a}} \mathrm{c}_{1}, \mathrm{c}_{2} \rightarrow \frac{1}{\mathrm{~b}} \mathrm{c}_{2}, \mathrm{c}_{3} \rightarrow \frac{1}{3} \mathrm{c}_{3}\right)\)
And applying \(\mathrm{R}_{1} \rightarrow \mathrm{R}_{1}+\mathrm{R}_{2}+\mathrm{R}_{3}\)
\(\left|\begin{array}{ccc}1+a^{2}+b^{2}+c^{2} & 1+a^{2}+b^{2}+c^{2} & 1+a^{2}+b^{2}+c^{2} \\ b^{2} & b^{2}+1 & b^{2} \\ c^{2} & c^{2} & c^{2}+1\end{array}\right|\)
\(=\left(1+a^{2}+b^{2}+c^{2}\right)\left|\begin{array}{ccc}1 & 0 & 0 \\ b^{2} & 1 & 0 \\ c^{2} & 0 & 1\end{array}\right|\)
on applying,
\(\left(c_{2} \rightarrow c_{2}-c_{1}, c_{3} \rightarrow c_{3}-c_{1}\right)\)
\(=\left(1+a^{2}+b^{2}+c^{2}\right) 1[1-0]\)
\(=\left(1+a^{2}+b^{2}+c^{2}\right)\)