Math needs at once?

∠QPR is 110°20' in the question, but 110°30' in the diagram. I suppose it is 110°30'.

In ΔVMP : MP = VM tan45° = VM
In ΔVMQ : MQ = VM tan45° = VM
In ΔVMP : MR = VM tan45° = VM
Hence, MP = MQ = MR
MP, MQ and MR are three radii of the circle with center M.

In a circle, ∠ at center = 2 (∠ in segment)
Reflex ∠QMR = 2 ∠QPR = 2 × 110.30° = 221°

∠s at a point : ∠QMR + (Reflex ∠QMR) = 360°
Hence, ∠QMR = 360° - 221° = 139°

In ΔMQR :
QR² = MQ² + MR² - 2 × MQ × MR × cos∠QMR (cosine law)

As shown above, MQ = MR = VM
QR² = VM² + VM² - 2 × VM² × cos139° (cosine law)
2VM² - 2 × VM² × cos139° = (6000 m)²
VM² (2 - 2 cos139°) = 6000² m²
VM = 6000 / √(2 - 2 cos139°) m
VM = 3203 m

The height of the mountain submit above the plane = 3203 m

(The given answer (1069 m) is incorrect.)
(If VM = 1609 m, then MQ + MR = 3218 m < QR.)
(This is impossible for ΔMQR.)