First, we need the equation of a line in the slope intercept form y = mx + b where m = gradient or slope, and b = y intercept.
In this case, our required line has a slope of 3, so y = 3x + b
This could be an infinity of line depending on value of b. But we need the specific value of b for the line which passes through (-6,1). So next step is to substitute the point values to get 1 = 3(-6) + b
=> b = 1 + 18 = 19
So our required line is y = 3x + 19
Finally, let's find the point (-4, r) on this line, by again substituting in the equation. This gives r = 3(-4) + 19
=> r = -12 + 19 = 7
Hence the required point is (-4, 7) lying on the line y = 3x + 19
You can double-check with Desmos Graphing or similar.