If the three planes $$x = 5$$, $$2x - 5ay + 3z - 2 = 0$$ and $$3bx + y - 3z = 0$$ contain a common line, then $$(a,b)$$ is equal to

The first plane is $$x = 5$$. Any point on the common line must therefore have $$x = 5$$.

Let a general point on the common line be $$(x,y,z)$$. Substituting $$x = 5$$ in the second plane

$$2x - 5ay + 3z - 2 = 0$$ gives
$$2(5) - 5a\,y + 3z - 2 = 0$$
$$10 - 5a\,y + 3z - 2 = 0$$
$$8 - 5a\,y + 3z = 0$$
$$3z = 5a\,y - 8$$
$$z = \frac{5a}{3}\,y - \frac{8}{3}\,.$$

Hence every point on the line can be written as
$$\bigl(x,\; y,\; z\bigr) = \Bigl(5,\; t,\; \frac{5a}{3}\,t - \frac{8}{3}\Bigr),$$ where $$t$$ is a real parameter.

This same point must satisfy the third plane
$$3bx + y - 3z = 0.$$

Substituting $$x = 5,\; y = t,\; z = \dfrac{5a}{3}t - \dfrac{8}{3}$$ gives
$$3b(5) + t - 3\!\left(\frac{5a}{3}t - \frac{8}{3}\right) = 0.$$

Simplify:
$$15b + t - 5a\,t + 8 = 0.$$

For this to hold for every value of the parameter $$t$$, both the coefficient of $$t$$ and the constant term must be zero:

1. Coefficient of $$t$$: $$1 - 5a = 0 \;\Longrightarrow\; a = \frac{1}{5}.$$

2. Constant term: $$15b + 8 = 0 \;\Longrightarrow\; b = -\frac{8}{15}.$$

Therefore$$(a,b) = \left(\frac{1}{5}, -\frac{8}{15}\right).$$

Option B which is: $$\left(\frac{1}{5}, -\frac{8}{15}\right)$$