Hello,

If two lines are parallel, then the slopes are equal.

The slope of y=6x+3 is 6.

Therefore, the equation of the line we want to find is y=6x+a, with a to determine.

Because it passes through (0,-10), we have -10=6*0+a. So a=-10.

Thus the equatio of the line is

y=ax+b

It passes through the point 0,-1. Therefore, -1=a*0+b --> b=-1

ax-y-1=0.

A normal vector to it is

.

A normal vector to 3x-2y+5=0 is

.

Because the two lines are perpendicular, their normal vector are perpendicular too. In particular, the scalar product is 0.

You have the equation of the line now;..

i hope i didn't make any mistake...