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**Hint:**If \[a\], \[b\] and \[c\] are the consecutive terms of an AP, then using the properties of arithmetic progression, \[b-a = c-b\].

Also, the area of the triangle with the vertices \[(x_{1},y_{1})\], \[(x_{2},y_{2})\] and \[(x_{3},y_{3})\] is given by the formula \[A = \dfrac{1}{2}[x_{1}(y_{2}-y_{3}) + x_{2}(y_{3}-y_{1}) + x_{3}(y_{1}-y_{2})]\].

**Complete step-by-step answer:**

Given the terms \[3k-2, 2k^{2}-5k+8\] and \[4k+3\] are the consecutive terms of an AP.

Using the properties of arithmetic progression, the difference between the consecutive terms are equal.

This implies,

\[2k^{2}-5k+8-(3k-2) = 4k+3-(2k^{2}-5k+8)\]

Solving them as follows:

\[\begin{align*}2k^{2}-5k+8-(3k-2) &= 4k+3-(2k^{2}-5k+8)\\ 2k^{2}-8k+10 &= -2k^{2}+9k-5\\

4k^{2}-17k+15 &= 0\\ 4k^{2}-12k-5k+15 &=0\\ 4k(k-3)+3(k-3) &=0\\ (k-3)(4k+3) &=0\\ k &=3, -\dfrac{3}{4}\end{align*}\]

So, the value of \[k = 3, -\dfrac{3}{4}\].

Now, the vertices of the triangle are \[(k,2)\], \[(3k,2)\] and \[(2,5)\], and its area is \[6\,\text{sq.u}\].

So, substituting the values into the formula for the area of the triangle, \[A = \dfrac{1}{2}[x_{1}(y_{2}-y_{3}) + x_{2}(y_{3}-y_{1}) + x_{3}(y_{1}-y_{2})]\], it gives,

\[\begin{align*}6 &= \dfrac{1}{2}[k(2-5)+3k(5-2)+2(2-2)]\\ 12 &= -3k+9k\\ 12 &= 6k\\ k&= 2\end{align*}\]

Therefore, the value of \[k\] is 2.

**Note:**The area of the triangle can also be evaluated using the lengths of the sides of the triangle, and then using Heron's formula to calculate the area.

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