OR is a side of the parallelogram OPQR, and since we have the coordinates of the endpoints, the equation of OR can be found. The slope of OR=5/3 and its equation is y=5x/3 because it passes through the origin O.
OP is parallel to QR so has the same slope. 3y+x=18 which can be written y=-⅓x+6, and so has the slope -⅓. The equation of OP is y=-⅓x, passing through O.
PQ is parallel to OR which has the slope 5/3 and its equation is y=5x/3+a where a is a constant and the y-intercept.
The question provides no information which would enable a to be found, and no picture has been provided. So I'm going to assume that the diagonal OQ is the y-axis (x=0). This implies that Q lies on the y-axis and is the y-intercept of 3y+x=18 and of y=5x/3+a, because, when x=0, 3y=18, y=6. Q therefore is the point (0,6). Since Q lies on the line PQ. The y-intercept of PQ is Q, so PQ is y=5x/3+6
P is the intersection of PQ and OP, so:
y=5x/3+6=-x/3,
6=-(5x+x)/3=-2x, making x=-3. So y=1 and P is (-3,1).
(a) Q(0,6)
(b) y=5x/3+6, or 3y-5x=18
(c) P(-3,1)