Matrices and Determinants Part 2

Given, M is a 3×3 matrix.
‌ Let ‌M=[

a	b	c
d	e	f
g	h	i

],‌ then ‌MT=[

a	d	g
b	e	h
c	f	i

]
Now, MTM=[

a	d	g
b	e	h
c	f	i

]×[

a	b	c
d	e	f
g	h	i

]
Sum of diagonal elements
=a2+b2+c2+d2+e2+f2+g2+h2+i2=7
According to the question, the entries are {0,1,2}.‌‌[∵M⊤M=7] i.e. {a,b,c,...,h,i}={0,1,2}
So, for Eq. (i) to be true, there are two cases.
Case I When 7−1′ s are there and 2−0 's are there.
⇒‌‌‌9C7×‌2C2=36 ways of arrangements.
Case II When 1−2 is there, 3−1′ s and 5−0′ s are there.
‌9C1×‌8C3×‌5C5=9×‌

8!

3!5!

×1
=504 ways of arrangements.
∴ Total possible arrangements =36+504=540