Express $(7^2 + 5^2 )(6^2 + 11^2 )(3^2 + 13^2 )(7^2 + 8^2 )$ in the form $(x^2 + y^2)(z^2 + w^2)$

Question

My friend comes up to me and says that you have 3 minutes to solve this question:
What does $(7^2 + 5^2 )(6^2 + 11^2 )(3^2 + 13^2 )(7^2 + 8^2 )$ equals to:

A) $(86^2 + 76^2)(130^2 + 29^2)$
B) $(97^2 + 47^2)(125^2 + 67^2)$
C) $(103^2 + 78^2)(47^2 + 88^2)$
D) $(9^2 + 91^2)(125^2 + 111^2)$
(Multiple correct options may exist!)

Now I thought, and thought why not brute force! It should work right? But then the clock told me no time for that. I cant think of anything else, Now I need your help with what other methods could I use to do this quickly. All unique ideas are welcome! Thanks
(I'm not asking for exact solution to the problem. Maybe a hint or two or some examples about the method you would apply to solve this)

EDIT: Thanks for pointing it out I put an extra 2 by mistake in all integers given in the question. It has been corrected now

Answer 1

HINT:

Half of the answer $\to $ Quickly check the last digit of each product.For example, the original product ends with 2: (a4)(b7)(c8)(d3)=(e2). You can go over the 4 options in similar way within a minute to eliminate (C) and (D), giving you 2 minutes to choose between (A) and (B).

Answer 2

Consider the complex number $Z = (7+5i)(7+8i)(6+11i)(3+13i)$

Simplifying, $Z=(49-40 +i(56+35))(18-143+i(78+33))$

$Z=(9+91i)(-125+111i)$

Compare the modulus now, you can work out similarly for the other possibilities.

I got the second expression for $Z$ by multiplying first and second independently from the product of third and fourth.