Given lines are:

3x + y = 4 … (i)

x – ay + 7 = 0 … (ii)

bx + 2y + 5 = 0 … (iii)

It’s said that these lines form three consecutive sides of a rectangle.

So,

Lines (i) and (ii) must be perpendicular

Also, lines (ii) and (iii) must be perpendicular

We know that, for two perpendicular lines the product of their slopes will be -1.

Now,

Slope of line (i) is

3x + y = 4 ⇒ y = -3x = 4

Hence, slope (m₁) = -3

And, slope of line (ii) is

x – ay + 7 = 0 ⇒ ay = x + 7

y = (1/a) x + 7/a

Hence, slope (m₂) = 1/a

Finally, the slope of line (iii) is

bx + 2y + 5 = 0 ⇒ 2y = -bx – 5

y = (-b/2) x – 5/2

Hence, slope (m₃) = -b/2

As lines (i), (ii) and (iii) are consecutive sides of rectangle, we have

m₁ x m₂ = -1 and m₂ x m₃ = -1

(-3) x (1/a) = -1 and (1/a) x (-b/2) = -1

-3 = -a and -b/2a = -1

a = 3 and b = 2a ⇒ b = 2(3) = 6

Thus, the value of a is 3 and the value of b is 6.