Let the parabola $$ y=x^{2}+px-3 $$, meet the coordinate axes at the points P, Q and R . If the circle C with centre at (-1, -1) passes through the points P, Q and R, then the area of $$ \triangle PQR $$ is :

The parabola is given by $$ y = x^2 + p x - 3 $$. It intersects the coordinate axes at points P, Q, and R.

To find the points of intersection:

When $$ x = 0 $$, $$ y = (0)^2 + p(0) - 3 = -3 $$. So, R is $$ (0, -3) $$.

When $$ y = 0 $$, $$ x^2 + p x - 3 = 0 $$. Let the roots be $$ x_1 $$ and $$ x_2 $$, so P and Q are $$ (x_1, 0) $$ and $$ (x_2, 0) $$.

The circle has center at $$ (-1, -1) $$ and passes through P, Q, and R. The radius is the distance from the center to any point on the circle.

Distance from center $$ (-1, -1) $$ to R $$ (0, -3) $$:

$$ \sqrt{ (0 - (-1))^2 + (-3 - (-1))^2 } = \sqrt{ (1)^2 + (-2)^2 } = \sqrt{1 + 4} = \sqrt{5} $$

So, the radius $$ r = \sqrt{5} $$.

Since the circle passes through P $$ (x_1, 0) $$ and Q $$ (x_2, 0) $$, the distance from the center to each must also be $$ \sqrt{5} $$.

For P $$ (x_1, 0) $$:

$$ \sqrt{ (x_1 - (-1))^2 + (0 - (-1))^2 } = \sqrt{ (x_1 + 1)^2 + 1^2 } = \sqrt{5} $$

Square both sides:

$$ (x_1 + 1)^2 + 1 = 5 $$

$$ (x_1 + 1)^2 = 4 $$

$$ x_1 + 1 = \pm 2 $$

So, $$ x_1 = 1 $$ or $$ x_1 = -3 $$.

Similarly, for Q $$ (x_2, 0) $$:

$$ (x_2 + 1)^2 + 1 = 5 $$

$$ (x_2 + 1)^2 = 4 $$

$$ x_2 = 1 \) or $$ x_2 = -3 $$.

Since the discriminant of $$ x^2 + p x - 3 = 0 $$ is $$ p^2 + 12 > 0 $$ (always positive), there are two distinct real roots. Therefore, the roots are 1 and -3. Assign P as $$ (1, 0) $$ and Q as $$ (-3, 0) $$.

Now, verify the value of p using the sum of roots:

Sum of roots: $$ x_1 + x_2 = 1 + (-3) = -2 = -p $$, so $$ p = 2 $$.

The parabola is $$ y = x^2 + 2x - 3 $$, and the points are P$$ (1, 0) $$, Q$$ (-3, 0) $$, and R$$ (0, -3) $$.

To find the area of $$ \triangle PQR $$ with vertices P$$ (1, 0) $$, Q$$ (-3, 0) $$, and R$$ (0, -3) $$, use the shoelace formula.

Shoelace formula for area given points $$ (x_1, y_1) $$, $$ (x_2, y_2) $$, $$ (x_3, y_3) $$:

$$ $$\text{Area} = \f\frac{1}{2}$$ $$\left$$| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) $$\right$$| $$

Substitute $$ x_1 = 1 $$, $$ y_1 = 0 $$, $$ x_2 = -3 $$, $$ y_2 = 0 $$, $$ x_3 = 0 $$, $$ y_3 = -3 $$:

$$ $$\text{Area} = \f\frac{1}{2}$$ $$\left$$| 1 $$\cdot$$ (0 - (-3)) + (-3) $$\cdot$$ (-3 - 0) + 0 $$\cdot$$ (0 - 0) $$\right$$| $$

$$ = $$\f\frac{1}{2}$$ $$\left$$| 1 $$\cdot$$ 3 + (-3) $$\cdot$$ (-3) + 0 $$\right$$| $$

$$ = $$\f\frac{1}{2}$$ $$\left$$| 3 + 9 $$\right$$| $$

$$ = $$\f\frac{1}{2}$$ $$\left$$| 12 $$\right$$| $$

$$ = $$\f\frac{1}{2}$$ $$\t\times$$ 12 = 6 $$

Therefore, the area of $$ \triangle PQR $$ is 6.