Let the distance of cliff from the initial position of man be ' d ' metres and speed of sound in air be ' V ' m∕ s.
For the first echo,
‌t1=3.5 s
‌V=‌

2d

t1

∴‌‌V=‌

2d

3⋅5

......(i)
On moving closer to the cliff by 84m, the distance of cliff from the new position becomes (d−84)m, then
for second echo, t2=3 s.
∵‌V‌=‌

2(d−84)

t2

∴‌V‌=‌

2(d−84)

3

.......(ii)
From (i) and (ii), we get
‌‌

2(d−84)

3

‌=‌

2d

3⋅5

⇒3⋅5(d−84)‌=3d
⇒3⋅5d−294‌=3d
⇒3⋅5d−3d‌=294
⇒‌0.5d‌=294
⇒‌d‌=‌

294

0.5

=588m.