The given line has intercepts (0,2) and (6,0) 
Here is the given line and the given point:

 

First we find the slope of the given line by getting it into the
slope-intercept form y = mx+b

x + 3y = 6
    3y = -x + 6
    3y = -1x + 6
    y = 

     y = x + 2

Compare that to

     y = mx + b

and the slope of the given line is m = .

Since the required line is to be perpendicular to the
given line, its slope is found by inverting the given
line's slope and giving it the opposite sign. Inverting
and changing the sign of  gives , or just 3.

Then we use the point-slope form:

y - y1 = m(x - x1)

with m = 3, (x1,y1) = (-3,-5).

y - (-5) = m[x - (-3)]

   y + 5 = 3(x + 3)

   y + 5 = 3x + 9
   
       y = 3x + 4

That's the required line.  Let's graph it as a check.
It has intercepts (0,4) and (,0), which is (,0), 
Let's draw it in green:


  
The green line whose equation we found sure looks perpendicular 
to the black one, whose equation was given, so it must be correct.

Answer:        y = 3x + 4

Edwin