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The problem involves a parallelogram ABCD with three given vertices A(-3, 2, x), B(3, -3, 1), and C(5, x, 2), and asks for the following:

a) **Determine the fourth vertex D.**
b) **Find the value of \( x \) such that the length of vector \( \overrightarrow{AD} \) equals \( \sqrt{5} \).**
c) **Calculate the angle between the diagonals of the parallelogram if \( x = -3 \).**

Let me break down the steps to solve these:

### Part a) Finding vertex D:
For a parallelogram, the midpoint of the diagonals coincide. Thus, the midpoint of diagonal \( AC \) must be the same as the midpoint of diagonal \( BD \).

- The midpoint of diagonal \( AC \) can be calculated as the average of the coordinates of A and C.
- Similarly, the midpoint of diagonal \( BD \) can be calculated as the average of the coordinates of B and D.

We will set these midpoints equal and solve for \( D \).

### Part b) Finding \( x \) so that \( \overrightarrow{AD} = \sqrt{5} \):
The vector \( \overrightarrow{AD} \) is the difference between the coordinates of \( A \) and \( D \). We need to set the length of this vector equal to \( \sqrt{5} \), and solve for \( x \).

### Part c) Calculating the angle between diagonals:
The angle between two vectors \( \overrightarrow{AC} \) and \( \overrightarrow{BD} \) can be found using the dot product formula:
\[
\cos(\theta) = \frac{\overrightarrow{AC} \cdot \overrightarrow{BD}}{|\overrightarrow{AC}| |\overrightarrow{BD}|}
\]
Here we calculate the vectors, take their dot product, and use the magnitude to find the angle.

Let me start solving this step by step.

#### Part a) Finding D:
We know the coordinates of A(-3, 2, x) and C(5, x, 2). The midpoint of AC is:

\[
M_{AC} = \left(\frac{-3 + 5}{2}, \frac{2 + x}{2}, \frac{x + 2}{2}\right) = \left(1, \frac{2 + x}{2}, \frac{x + 2}{2}\right)
\]

Similarly, the midpoint of BD is:

\[
M_{BD} = \left(\frac{3 + x_1}{2}, \frac{-3 + y_1}{2}, \frac{1 + z_1}{2}\right)
\]

Let me proceed to solve these to determine D.  sympy import symbols, Eq, solve

# Define symbols for x1, y1, z1 (coordinates of D)
x1, y1, z1 = symbols('x1 y1 z1')

# Coordinates of A and C
A = (-3, 2, 'x')  # A = (-3, 2, x)
C = (5, 'x', 2)   # C = (5, x, 2)

# Midpoint of AC = M_AC = ( (A_x + C_x) / 2, (A_y + C_y) / 2, (A_z + C_z) / 2 )
M_AC = ((A[0] + C[0]) / 2, (A[1] + C[1]) / 2, (A[2] + C[2]) / 2)

# Coordinates of B
B = (3, -3, 1)  # B = (3, -3, 1)

# Midpoint of BD = M_BD = ((B_x + D_x) / 2, (B_y + D_y) / 2, (B_z + D_z) / 2 )
M_BD = ((B[0] + x1) / 2, (B[1] + y1) / 2, (B[2] + z1) / 2)

# Set midpoint of AC equal to midpoint of BD
eq1 = Eq(M_AC[0], M_BD[0])
eq2 = Eq(M_AC[1], M_BD[1])
eq3 = Eq(M_AC[2], M_BD[2])

# Solve for x1, y1, z1 (coordinates of D)
solution = solve((eq1, eq2, eq3), (x1, y1, z1))
solutionIt seems that there was an error because of the presence of the variable \( x \) in the coordinates of A and C. I will correct the approach and ensure that the variable \( x \

Dana so tri oglišča paralelograma ABCD: A(-3,2,x), B(3,-3,1) in C(5,x,2). a) Določi četrto oglišče D. b) Določi parameter x tako, da bo dolžina vektorja AD enaka √5. c) Izračunaj kot med diagonalama paralelograma, če je x = -3.

Solve a coordinate geometry problem involving parallelogram ABCD with vertices A(-3,2,x), B(3,-3,1), and C(5,x,2). Learn how to find the fourth vertex D, determine parameter x for a specific vector length, and calculate the angle between diagonals when x = -3 using vector and midpoint formulas.

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