Let m be the slope of the line,
then tan60o=∣1−m3m+3∣
⇒3=∣1−3mm+3∣
⇒±3=1−3mm+3
∴3−3m=m+3 or −3+3m=m+3
4m=0 or 2m=23
m=0 or m=3
Since L intersects X-axis we take m=3
and get y=3x+c
Since it passes through (3,−2), therefore line will become y−3x+2+33=0