Calculation of $\lambda$ in determinant multiplication. $$\begin{vmatrix}
a^2+\lambda^2 & ab+c\lambda & ca-b\lambda \\ 
ab-c\lambda &  b^2+\lambda^2& bc+a\lambda\\ 
ca+b\lambda & bc-a\lambda & c^2+\lambda^2
\end{vmatrix}.\begin{vmatrix}
\lambda & c & -b\\ 
 -c& \lambda & a\\ 
b & -a & \lambda
\end{vmatrix}=(1+a^2+b^2+c^2)^3.$$
Then value of $\lambda$ is 
options:: (a)$\;  8\;\;$    (b) $\;  27\;\;$ (c)$ \;  1\;\;$ (d) $\;  -1\;\;$
actually as I have seen the question. Then I used Multiply these two determinant. but This is very tidious task.
So I want a better Method by which we can easily calculate value of $\lambda$.
So please explain me in detail.
 A: I am not sure this is the best way, but brute force leads to:
$$\lambda^3(\lambda^2 + a^2 + b^2 + c^2)^3 = (1+a^2+b^2+c^2)^3$$
It is very clear what $\lambda$ has to be.
The first determinant yields: $\lambda^2(a^2 + b^2 + c^2 + \lambda^2)^2.$
The second determinant yields: $\lambda(a^2 + b^2 + c^2 + \lambda^2).$
Maybe there is an easy way to take advantage of the second determinant with the RHS, but I am tired and do not see it.
A: Since this is a multiple choice question, and the two sides are meant to be equal as functions of $a,b,c$, it remains true when you substitute $a=b=c=0$, giving $\lambda^9 = 1$.
A: Note: You are given an equation, which can be viewed as the equivalence of two functions of $a, b, c$: $f(a, b, c) = g(a, b, c)$, and the equivalence will hold no matter what we input for $(a, b, c)$, so long as the input of each function is the same. 
So let's input $(a, b, c) = (0, 0, 0)$, simplify $f(0, 0, 0) = g(0, 0, 0)$, and solve for $\lambda$.
$$\begin{vmatrix}
a^2+\lambda^2 & ab+c\lambda & ca-b\lambda \\ 
ab-c\lambda &  b^2+\lambda^2& bc+a\lambda\\ 
ca+b\lambda & bc-a\lambda & c^2+\lambda^2
\end{vmatrix}.\begin{vmatrix}
\lambda & c & -b\\ 
 -c& \lambda & a\\ 
b & -a & \lambda
\end{vmatrix}=(1+a^2+b^2+c^2)^3.$$
$$
\implies \begin{vmatrix}
\lambda^2 & 0 & 0 \\ 
0 &  \lambda^2& 0\\ 
0 & 0 & \lambda^2
\end{vmatrix}.\begin{vmatrix}
\lambda & 0 & 0\\ 
 0& \lambda & 0\\ 
0 & 0 & \lambda
\end{vmatrix}=(1)^3.$$ 
$$ \iff \lambda^6\cdot \lambda^3 = 1 \iff \lambda^9 = 1 \iff \lambda = 1.$$ 

Comment
Since the question is multiple choice, we can feel confident that option $(c)$ is the correct value for $\lambda$. If you had been asked to prove or disprove the equivalence, then you'd have a complicated mess to work out. But given the equivalence, and being asked to determine that given the equivalence, "what must be the value of $\lambda$?" --> you can use that equivalence to your advantage, since it must hold for all possible triplets $(a, b, c)$, we use the triplet that makes for the least work.
A: Hints: remember the multilinearity of the determinant:
$$\begin{vmatrix}
a^2+\lambda^2 & ab+c\lambda & ca-b\lambda \\ 
ab-c\lambda &  b^2+\lambda^2& bc+a\lambda\\ 
ca+b\lambda & bc-a\lambda & c^2+\lambda^2
\end{vmatrix}=\begin{vmatrix}
a^2 & ab& ca \\ 
ab &  b^2& bc\\ 
ca& bc& c^2
\end{vmatrix}+\begin{vmatrix}
a^2 & ab& -b\lambda \\ 
ab &  b^2& a\lambda\\ 
ca& bc& \lambda^2
\end{vmatrix}+\ldots$$
So, for example, the first two determinants explicityly written down above are zero , since in the first one $\,a=0\,$ or the second line of the first is the first one divided by $\,a\,$ and then multiplied by $\,b\,$ , and in the second one the second column...etc. Develop fully the above to simplify your calculations
A: Compute some cofactors of the second matrix.  Compare with corresponding elements of the first matrix.  This should lead to a solution without a lot of brute force.
Added:  What I had in mind was the following.  Let
$$M_1=\begin{bmatrix}
a^2+\lambda^2 & ab+c\lambda & ca-b\lambda \\ 
ab-c\lambda &  b^2+\lambda^2& bc+a\lambda\\ 
ca+b\lambda & bc-a\lambda & c^2+\lambda^2
\end{bmatrix}$$
and let
$$M_2=\begin{bmatrix}
\lambda & c & -b\\ 
 -c& \lambda & a\\ 
b & -a & \lambda
\end{bmatrix}$$
Then the equation amounts to $\det M_1\cdot \det M_2=(1+a^2+b^2+c^2)^3.$  It's quick to compute the cofactors of $M_2$ since they're all $2\times 2$ determinants.  You'll find that $M_1$ is the cofactor matrix of $M_2.$
So $M_1^TM_2=(\det M_2)I$ and therefore
$$\det M_1\cdot\det M_2=\det(M_1^TM_2)=\det((\det M_2)I)=(\det M_2)^3.$$
The determinant of $M_2$ is also quick to compute; it's $\lambda(\lambda^2+a^2+b^2+c^2).$  Now equate the cube of this expression to $(1+a^2+b^2+c^2)^3.$
