(2) $$ A B C D $$ is a parallelogram such that $$ A \subset(5,2) $$, $$ B(-2,-3) $$ and $$ \overrightarrow{A C}=\binom{1}{3} $$ Findli.) the coorctinates of Cand $$ D \cdot \overrightarrow{B C} $$

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**Step 1. Find the coordinates of $$ C $$.**

We are given that  
$$
\overrightarrow{A C} = \begin{pmatrix} 1 \ 3 \end{pmatrix}.
$$
Since $$ \overrightarrow{A C} = C - A $$, we have  
$$
C = A + \overrightarrow{A C}.
$$
Given $$ A = (5,2) $$, we obtain  
$$
C = (5,2) + (1,3) = (5+1,\, 2+3) = (6,5).
$$

**Step 2. Find the coordinates of $$ D $$.**

In a parallelogram, one useful property is that the diagonals bisect each other. The midpoint of diagonal $$ AC $$ is  
$$
M = \left(\frac{5+6}{2},\, \frac{2+5}{2}\right) = \left(\frac{11}{2},\, \frac{7}{2}\right).
$$
Similarly, the midpoint of diagonal $$ BD $$, with $$ B = (-2,-3) $$ and unknown $$ D $$, must also equal $$ M $$. Let $$ D = (x_D, y_D) $$. Then,
$$
\left(\frac{-2+x_D}{2},\, \frac{-3+y_D}{2}\right) = \left(\frac{11}{2},\, \frac{7}{2}\right).
$$
Equate the components:
$$
\frac{-2+x_D}{2} = \frac{11}{2} \quad \Longrightarrow \quad -2+x_D=11 \quad \Longrightarrow \quad x_D=13,
$$
$$
\frac{-3+y_D}{2} = \frac{7}{2} \quad \Longrightarrow \quad -3+y_D=7 \quad \Longrightarrow \quad y_D=10.
$$
Thus,  
$$
D=(13,10).
$$

**Step 3. Find $$ \overrightarrow{B C} $$.**

The vector $$ \overrightarrow{B C} $$ is given by  
$$
\overrightarrow{B C}=C-B.
$$
With $$ C=(6,5) $$ and $$ B=(-2,-3) $$, we have  
$$
\overrightarrow{B C}=(6-(-2),\, 5-(-3))=(8,8).
$$

**Final Answers:**

- $$ C = (6,5) $$
- $$ D = (13,10) $$
- $$ \overrightarrow{B C} = (8,8) $$
$$ C = (6,5) $$, $$ D = (13,10) $$, and $$ \overrightarrow{B C} = (8,8) $$.

(2) is a parallelogram such that ,
and
Findli.) the coorctinates of Cand

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(2) is a parallelogram such that , and Findli.) the coorctinates of Cand