Let the point $$P$$ be the vertex of the parabola $$y = x^2 - 6x + 12$$. If a line passing through the point $$P$$ intersects the circle $$x^2 + y^2 - 2x - 4y + 3 = 0$$ at the points $$R$$ and $$S$$.then the maximum value of $$(PR + PS)^2$$ is :

The vertex form of the parabola $$y = x^2 - 6x + 12$$ is obtained by completing the square:

$$y = (x-3)^2 + 3$$

Hence the vertex is $$P(3,\,3)$$.

Rewrite the circle $$x^2 + y^2 - 2x - 4y + 3 = 0$$ in centre-radius form:

$$x^2 - 2x + y^2 - 4y = -3$$
$$(x-1)^2 - 1 + (y-2)^2 - 4 = -3$$
$$(x-1)^2 + (y-2)^2 = 2$$

So the circle has centre $$C(1,\,2)$$ and radius $$r=\sqrt2$$.

Distance of $$P$$ from the centre:

$$PC = \sqrt{(3-1)^2 + (3-2)^2} = \sqrt{4+1} = \sqrt5$$

The power of point $$P$$ with respect to the circle is

$$\text{Pow}(P) = PC^2 - r^2 = 5 - 2 = 3$$

For every line through an external point, the product of the lengths of the intercepted segments equals the power. Hence for any secant $$PRS$$ through $$P$$,

$$PR \cdot PS = 3 \quad -(1)$$

Let the direction cosines of the line be $$\cos\theta = c$$ and $$\sin\theta = s$$ (so $$c^2+s^2=1$$). Parametric form of the line through $$P(3,3)$$ is

$$x = 3 + tc,\;\; y = 3 + ts$$

Substituting in the circle equation:

$$(3+tc)^2 + (3+ts)^2 - 2(3+tc) - 4(3+ts) + 3 = 0$$
$$t^2 + 2(2c+s)\,t + 3 = 0$$

Let the roots be $$t_1, t_2$$ (with $$t_1, t_2 \gt 0$$ because $$P$$ is outside the circle and both intersection points lie ahead along a suitable direction). Then

$$t_1 + t_2 = -2(2c+s), \qquad t_1t_2 = 3 \quad\bigl(\text{matches }(1)\bigr)$$

The required quantity is

$$PR + PS = t_1 + t_2 = -2(2c+s)$$

Since we need a positive length, choose directions where $$2c+s \lt 0$$; then

$$PR + PS = 2\,\bigl|2c+s\bigr|$$

To maximise $$\lvert 2c+s\rvert$$ subject to $$c^2+s^2=1$$, recognise that $$2c+s$$ is the dot‐product of the unit vector $$(c,s)$$ with the fixed vector $$(2,1)$$. The extremal value of a dot‐product equals the product of the magnitudes:

$$\max\bigl|2c+s\bigr| = \sqrt{2^2 + 1^2} = \sqrt5$$

(This maximum is attained when $$(c,s)$$ is antiparallel to $$(2,1)$$, i.e. in the direction of $$(-2,-1)$$.)

Therefore

$$\bigl(PR + PS\bigr)_{\max} = 2\sqrt5$$

and

$$(PR + PS)_{\max}^2 = (2\sqrt5)^2 = 20$$

Option B which is: $$20$$