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