In this problem, no, you can never have a velocity of $\displaystyle v = \sqrt{61}$ because the least velocity it can attain is when the vertical velocity is zero and only the horizontal velocity is present, that is a velocity of 10 m/s, and this occurs at the highest point reached. And at a different velocity, you will have a different angle, hence a different distance the projectile will cover. Everything changes when you change either the value of the initial velocity, its angle of projection, or the acceleration along the vertical. Hence why there is only one possible vertical and horizontal component.

Which 3 lines are you referring to?

You could say that... but it's more complicated. It is a position graph at multiple times.

At time 0, the projectile is at (0, 0)

At time 1, the projectile is at (10, 11)

And so on.

First, a vector consists of

__a__ magnitude (that you call 'value' here) and then it has

__a__ direction. A graph is often used to describe a vector. If you are comparing a line with a vector, like the line y = x with a vector, they are different things.