JEE Main 2023 — Mathematics Algebra

Let

$${x}^{p{q}^{2}}={y}^{qr}={z}^{{p}^{2}r}=\lambda$$xpq2=yqr=zp2r=λ

Then,

$$p{q}^{2}={\mathrm{log}}_{x}\lambda$$pq2=logx​λ

$$qr={\mathrm{log}}_{y}\lambda$$qr=logy​λ

$${p}^{2}r={\mathrm{log}}_{z}\lambda$$p2r=logz​λ

So,

$$\frac{{\mathrm{log}}_{y}\lambda }{{\mathrm{log}}_{x}\lambda }=\frac{qr}{p{q}^{2}}$$logx​λlogy​λ​=pq2qr​

$$\Rightarrow {\mathrm{log}}_{y}x=\frac{r}{pq}...(1)$$⇒logy​x=pqr​...(1)

Similarly,

$${\mathrm{log}}_{x}z=\frac{p{q}^{2}}{{p}^{2}r}=\frac{{q}^{2}}{pr}....(2)$$logx​z=p2rpq2​=prq2​....(2)

$${\mathrm{log}}_{z}y=\frac{{p}^{2}r}{qr}=\frac{{p}^{2}}{q}...(3)$$logz​y=qrp2r​=qp2​...(3)

So, $3,3{\mathrm{log}}_{y}x,3{\mathrm{log}}_{z}y,7{\mathrm{log}}_{x}z$3,3logy​x,3logz​y,7logx​z are in A.P., hence $3,\frac{3r}{pq},\frac{3{p}^{2}}{q},\frac{7{q}^{2}}{pr}$3,pq3r​,q3p2​,pr7q2​ are in A.P. Therefore,

$$\frac{3r}{pq}-3=\frac{1}{2}$$pq3r​−3=21​

$$r=\frac{7}{6}pq$$r=67​pq

Also,

$$r=pq+1$$r=pq+1

So,

$$pq+1=\frac{7}{6}pq$$pq+1=67​pq

$$\Rightarrow pq=6$$⇒pq=6

$$\Rightarrow r=7$$⇒r=7

So,

$$\frac{3{p}^{2}}{q}=3+2(\frac{1}{2})=4$$q3p2​=3+2(21​)=4

$$\Rightarrow \frac{3{p}^{3}}{pq}=4$$⇒pq3p3​=4

$$\Rightarrow {p}^{3}=8$$⇒p3=8

$$\Rightarrow p=2$$⇒p=2

$$\Rightarrow q=3$$⇒q=3

Hence, $r-p-q=7-2-3=2$r−p−q=7−2−3=2