We are given the following line and circle equations:

1. Line equation: $3x - y - 9 = 0$
2. Circle equation: $x^2 + px + y^2 + 4y = 20$

Step 1: Express $y$ from the line equation.

From the line equation $3x - y - 9 = 0$, we can solve for $y$:

$y = 3x - 9$

Step 2: Substitute $y = 3x - 9$ into the circle equation.

Substitute this expression for $y$ into the circle equation $x^2 + px + y^2 + 4y = 20$:

$x^2 + px + (3x - 9)^2 + 4(3x - 9) = 20$

Expand $(3x - 9)^2$ and $4(3x - 9)$:

$(3x - 9)^2 = 9x^2 - 54x + 81$ $4(3x - 9) = 12x - 36$

Thus, the circle equation becomes:

$x^2 + px + 9x^2 - 54x + 81 + 12x - 36 = 20$

Simplify the equation:

$10x^2 + (p - 42)x + 45 = 0$

Step 3: Condition for no intersection.

For the line not to intersect with the circle, the quadratic equation $10x^2 + (p - 42)x + 45 = 0$ must have no real solutions. This occurs when the discriminant $\Delta$ is less than 0. The discriminant of a quadratic equation $ax^2 + bx + c = 0$ is given by:

$\Delta = b^2 - 4ac$

Here, $a = 10$, $b = p - 42$, and $c = 45$. So, the discriminant is:

$\Delta = (p - 42)^2 - 4(10)(45)$

Simplify:

$\Delta = (p - 42)^2 - 1800$

For no intersection, we require $\Delta < 0$:

$(p - 42)^2 < 1800$

Taking the square root of both sides:

$|p - 42| < \sqrt{1800}$

$|p - 42| < 42.426$

Thus, the range for $p$ is:

$42 - \sqrt{1800} < p < 42 + \sqrt{1800}$

Since $\sqrt{1800} = \sqrt{10 \times 180} = \sqrt{10} \times \sqrt{180} \approx 42.426$, this gives us the range:

$42 - \sqrt{10} < p < 42 + \sqrt{10}$

This shows that the value $p$ lies within that