In the diagram, C is the point (30,15), A is a point on the y-axis and B is a point on the x-axis.

1. Find the coordinates of the endpoints $A$ and $B$ from the line equation.

Since $A$ is on the $y$-axis, $x=0$.

$3y+0=15 \Rightarrow y=5$, so $A=(0,5)$.

Since $B$ is on the $x$-axis, $y=0$.

$0+x=15 \Rightarrow x=15$, so $B=(15,0)$.

2. Calculate the distance $AB$.

$AB=\sqrt{(15-0)^2+(0-5)^2}$
$=\sqrt{225+25}$
$=\sqrt{250}=5\sqrt{10}$

1. Find the gradient of line $AB$ from its equation.

Given line $AB$:
$3y+x=15$
$3y=-x+15$
$y=-\frac{1}{3}x+5$
So the gradient of $AB$ is $-\frac{1}{3}$.

A line parallel to $AB$ has the same gradient, so its gradient is also $-\frac{1}{3}$.

2. Use the point $C(30,15)$ to find the equation of the parallel line.

Using $y=mx+c$ with $m=-\frac{1}{3}$ and passing through $C(30,15)$:
$15=-\frac{1}{3}(30)+c$
$15=-10+c$
$c=25$
Hence,
$y=-\frac{1}{3}x+25$
Multiplying by 3:
$3y=-x+75$
$3y+x=75$

1. Use the parallelogram property to find $D$ from $A$, $B$ and $C$.

From the line $3y+x=15$, the intercepts are
$A=(0,5)$ and $B=(15,0)$.

Since $ABCD$ is a parallelogram,
$\overrightarrow{AD}=\overrightarrow{BC}.$
$\overrightarrow{BC}=(30-15,\,15-0)=(15,15).$
Hence
$D=A+\overrightarrow{BC}=(0,5)+(15,15)=(15,20).$