Calculation:

Since, plane of \(\displaystyle\triangle{A}'{B}'{C}'\) is similar to the plane of \(\displaystyle\triangle{A}{B}{C}\)

Since, ABCD is a parallelogram.

\(\displaystyle\angle{B}=\angle{V}\therefore\)(Common)

\(\displaystyle\angle{V}{A}{B}=\angle{V}{A}'{B}'\therefore\)(Corresponding angle in two similar plane)

Therefore by AA similarity, \(\displaystyle\triangle{V}{A}{B}\sim\triangle{V}{A}'{B}'\)

Ratio of corresponding sides in two similar triangles is equal.

\(\displaystyle\frac{{{V}{A}}}{{{V}{A}'}}=\frac{{{V}{C}}}{{{V}{C}'}}\)

\(\displaystyle\frac{{{V}{A}'+\forall'}}{{{V}{A}'}}=\frac{{{V}{C}}}{{{V}{C}'}}\)

\(\displaystyle\frac{{{15}+{20}}}{{{15}}}=\frac{{{V}{C}}}{{{18}}}\)

\(\displaystyle{V}{C}=\frac{{{18}\times{35}}}{{{15}}}\)

VC=42

Therefore, the answer is 42.

Since, plane of \(\displaystyle\triangle{A}'{B}'{C}'\) is similar to the plane of \(\displaystyle\triangle{A}{B}{C}\)

Since, ABCD is a parallelogram.

\(\displaystyle\angle{B}=\angle{V}\therefore\)(Common)

\(\displaystyle\angle{V}{A}{B}=\angle{V}{A}'{B}'\therefore\)(Corresponding angle in two similar plane)

Therefore by AA similarity, \(\displaystyle\triangle{V}{A}{B}\sim\triangle{V}{A}'{B}'\)

Ratio of corresponding sides in two similar triangles is equal.

\(\displaystyle\frac{{{V}{A}}}{{{V}{A}'}}=\frac{{{V}{C}}}{{{V}{C}'}}\)

\(\displaystyle\frac{{{V}{A}'+\forall'}}{{{V}{A}'}}=\frac{{{V}{C}}}{{{V}{C}'}}\)

\(\displaystyle\frac{{{15}+{20}}}{{{15}}}=\frac{{{V}{C}}}{{{18}}}\)

\(\displaystyle{V}{C}=\frac{{{18}\times{35}}}{{{15}}}\)

VC=42

Therefore, the answer is 42.