Matrices and Determinants Part 2

Method (I)
Given, x,y and z are in AP with common difference =d
∴‌‌x= First term
y= Second term of AP = First term + Common difference
⇒‌‌y=x+d...(i)
and z= Third term of AP = Second term + Common difference
⇒‌‌z=(x+d)+d=x+2d...(ii)
Also, given x≠3d.....(iii)
and [

3	4√2	x
4	5√2	y
5	k	z

]=0
Applying R2→R1+R3−2R2, we have
|

3	4√2	x
0	k−6√2	0
5	k	z

|=0
⇒(k−6√2)(3z−5x)=0( Expanding along R2)
Either k−6√2=0 or 3z−5x=0
⇒ k=6√2 or 3(x+2d)−5x=0  [from Eq. (ii)]
⇒x=3d which is not possible as in Eq. (iii).
∴‌‌k=6√2 is only one solution.
Hence, k2=(6√2)2
⇒‌‌k2=72