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Answer: $4 a^{2} b^{2} c^{2}$

Hint: Use determinant formula

Given: $\left|\begin{array}{ccc} b^{2}+c^{2} & a b & a c \\ b a & c^{2}+a^{2} & b c \\ c a & c b & a^{2}+b^{2} \end{array}\right|=4 a^{2} b^{2} c^{2}$

Solution:

\begin{aligned} &\text { L.H.S }\left|\begin{array}{ccc} b^{2}+c^{2} & a b & a c \\ b a & c^{2}+a^{2} & b c \\ c a & c b & a^{2}+b^{2} \end{array}\right| \\ &\text { Use } R_{1} \rightarrow a R_{1}, R_{2} \rightarrow b R_{2}, R_{3} \rightarrow c R_{3} \\ &=\frac{1}{a b c}\left|\begin{array}{ccc} a\left(b^{2}+c^{2}\right) & a^{2} b & a^{2} c \\ b^{2} a & b\left(c^{2}+a^{2}\right) & b^{2} c \\ c^{2} a & c^{2} b & c\left(a^{2}+b^{2}\right) \end{array}\right| \end{aligned}

\begin{aligned} &\text { Take common a from } \mathrm{R}_{1}, b \text { from } \mathrm{R}_{2} \text { and } \mathrm{c} \text { from } \mathrm{R}_{3}\\ &=\frac{a b c}{a b c}\left|\begin{array}{ccc} \left(b^{2}+c^{2}\right) & a^{2} & a^{2} \\ b^{2} & \left(c^{2}+a^{2}\right) & b^{2} \\ c^{2} & c^{2} & \left(a^{2}+b^{2}\right) \end{array}\right| \end{aligned}

\begin{aligned} &\text { Use } R_{1} \rightarrow R_{1}-R_{2}-R_{3} \\ &=\left|\begin{array}{ccc} 0 & -2 c^{2} & -2 b^{2} \\ b^{2} & \left(c^{2}+a^{2}\right) & b^{2} \\ c^{2} & c^{2} & \left(a^{2}+b^{2}\right) \end{array}\right| \end{aligned}

\begin{aligned} &\text { Expanding w.r.t } C_{1}\\ &=-b^{2}\left|\begin{array}{cc} -2 c^{2} & -2 b^{2} \\ c^{2} & a^{2}+b^{2} \end{array}\right|+c^{2}\left|\begin{array}{cc} -2 c^{2} & -2 b^{2} \\ c^{2}+a^{2} & b^{2} \end{array}\right|\\ &=-b^{2}\left(-2 a^{2} c^{2}-2 b^{2} c^{2}+2 b^{2} c^{2}\right)+c^{2}\left(-2 c^{2} b^{2}+2 b^{2} c^{2}+2 a^{2} b^{2}\right)\\ &=2 a^{2} b^{2} c^{2}+2 a^{2} b^{2} c^{2}=4 a^{2} b^{2} c^{2}\\ &=R . H . S \end{aligned}

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