Given:

The equation is perpendicular to x – 3y + 5 = 0 and passes through (3,-2)

The equation of the line perpendicular to x − 3y + 5 = 0 is

3x + y + λ = 0,

Where, λ is a constant.

It passes through (3, − 2).

Substitute the values in above equation, we get

3 (3) + (-2) + λ = 0

9 – 2 + λ = 0

λ = – 7

Now, substitute the value of λ = − 7 in 3x + y + λ = 0, we get

3x + y – 7 = 0

The required line is 3x + y – 7 = 0.

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