Consider the following schedule:
S = $(2, W, x), (1, W, z), (1, W, x), C_{1}, (2, W, y), (2, R, y), (3, W, z)$

$w h e r e C_{1} means commit of 1st transaction.$
Triple(i, W/R,x) means transaction i writes/reads item x,

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Solution

Check for View serializablity:
Transaction 2 is reading item m which is written by trans 3, hence, In the view equivalent serial schedule, trans 2 must appear after (not necessarily immediately) trans 3. (3 $\to$ 2)
Transaction 1 is doing final write of item x which is also written by trans 2, hence, in the view equivalent serial schedule, trans 1 must appear after (not necessarily immediately) trans 2.
(2 $\to$ 1) Transaction 3 is doing final write of item z which is also written by trans 1, hence, in the view equivalent serial schedule, trans 3 must appear after (not necessarily immediately) Trans 1. (1 $\to$ 3) But now we have a contradiction. 3 $\to$ 2 $\to$ 1 but also 1 $\to$ 3.
Hence, not view serializable.

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