Calculate the value of a so that the vectors (2,2,−1),(3,4,2) and(a,2,3) are coplanar.

To determine the value of "a" such that the given vectors are coplanar, we can use the fact that three vectors are coplanar if and only if their determinant is equal to zero.

So, let's set up the determinant using the given vectors:

\begin{vmatrix}2 & 2 & -1 \3 & 4 & 2 \a & 2 & 3 \\end{vmatrix}

Expanding the determinant along the top row, we get:

2 \begin{vmatrix}4 & 2 \2 & 3 \\end{vmatrix}- 2 \begin{vmatrix}3 & 2 \a & 3 \\end{vmatrix}- 1 \begin{vmatrix}3 & 4 \a & 2 \\end{vmatrix}

Calculating the determinants:

2((4)(3) - (2)(2))- 2((3)(3) - (2)(a))- 1((3)(2) - (4)(a))

Simplifying:

2(12 - 4) - 2(9 - 2a) - 1(6 - 4a)

24 - 8 - 18 + 4a - 6 + 4a

4a + 4a - 8 - 18 - 6

8a - 32

Now, we want the determinant to be equal to zero:

8a - 32 = 0

Solving for "a":

8a = 32

a = \frac{{32}}{{8}} = 4

Answer: The value of "a" that makes the vectors (2,2,−1), (3,4,2), and (a,2,3) coplanar is \boxed{4}.