By using properties of determinants, show that:

$\left|\begin{array}{ccc}a^{2}+1 & a b & a c \\ a b & b^{2}+1 & b c \\ c a & c b & c^{2}+1\end{array}\right|=1+a^{2}+b^{2}+c^{2}$

$\Delta=\left|\begin{array}{ccc}a^{2}+1 & a b & a c \\ a b & b^{2}+1 & b c \\ c a & c b & c^{2}+1\end{array}\right|$

Taking out common factors $a, b$ and $c$ from $R_{1}, R_{2}$ and $R_{3}$ respectively, we have:

$\Delta=a b c\left|\begin{array}{ccc}a+\frac{1}{a} & b & c \\ a & b+\frac{1}{b} & c \\ a & b & c+\frac{1}{c}\end{array}\right|$

Applying $R_{2} \rightarrow R_{2}-R_{1}$ and $R_{3} \rightarrow R_{3}-R_{1},$ we have:

$\Delta=a b c\left|\begin{array}{ccc}a+\frac{1}{a} & b & c \\ -\frac{1}{a} & \frac{1}{b} & 0 \\ -\frac{1}{a} & 0 & \frac{1}{c}\end{array}\right|$

Applying $C_{1} \rightarrow a C_{1}, C_{2} \rightarrow b C_{2}$ and $C_{3} \rightarrow c C_{3},$ we have:

$\Delta=a b c \times \frac{1}{a b c}\left|\begin{array}{ccc}a^{2}+1 & b^{2} & c^{2} \\ -1 & 1 & 0 \\ -1 & 0 & 1\end{array}\right|$

Expanding along $R_{3},$ we have:

$\Delta=-1\left|\begin{array}{ll}b^{2} & c^{2} \\ 1 & 0\end{array}\right|+1\left|\begin{array}{ll}a^{2}+1 & b^{2} \\ -1 & 1\end{array}\right|$

$=-1\left(-c^{2}\right)+\left(a^{2}+1+b^{2}\right)=1+a^{2}+b^{2}+c^{2}$

Hence, the given result is proved.