Consider the lines $$x(3\lambda + 1) + y(7\lambda + 2) = 17\lambda + 5$$, $$\lambda$$ being a parameter, all passing through a point P. One of these lines (say L) is farthest from the origin. If the distance of L from the point $$(3, 6)$$ is d, then the value of $$d^2$$ is

The family of lines is given by $$(3\lambda + 1)x + (7\lambda + 2)y = 17\lambda + 5$$, where $$\lambda$$ is a real parameter.

Step 1 : Find the fixed point $$P(x_0,y_0)$$
Since every member of the family passes through the same point $$P$$, substitute $$x = x_0,\; y = y_0$$ and demand that the resulting identity hold for all $$\lambda$$:

$$(3\lambda + 1)x_0 + (7\lambda + 2)y_0 - (17\lambda + 5) = 0 \quad \text{for all }\lambda.$$

Equating the coefficients of $$\lambda$$ and the constant terms separately, we get
$$3x_0 + 7y_0 - 17 = 0 \;-(1)$$
$$x_0 + 2y_0 - 5 = 0 \;-(2)$$

From $$(2)$$, $$x_0 = 5 - 2y_0$$. Substitute into $$(1)$$:
$$3(5 - 2y_0) + 7y_0 - 17 = 0$$
$$15 - 6y_0 + 7y_0 - 17 = 0$$
$$y_0 - 2 = 0 \;\Rightarrow\; y_0 = 2$$
$$x_0 = 5 - 2(2) = 1$$

Hence $$P(1,2)$$ is the common point.

Step 2 : Distance of a general member from the origin
Write the line in the form $$Ax + By - C = 0$$:
$$A = 3\lambda + 1,\; B = 7\lambda + 2,\; C = 17\lambda + 5.$$
The perpendicular distance from the origin $$(0,0)$$ is
$$D(\lambda) = \frac{|C|}{\sqrt{A^2 + B^2}} = \frac{|17\lambda + 5|} {\sqrt{(3\lambda + 1)^2 + (7\lambda + 2)^2}}.$$

To locate the farthest line from the origin, maximise $$D(\lambda)$$. It is more convenient to maximise its square:

$$f(\lambda) = \frac{(17\lambda + 5)^2} {(3\lambda + 1)^2 + (7\lambda + 2)^2}.$$

Denominator simplification:
$$(3\lambda + 1)^2 + (7\lambda + 2)^2 = (9\lambda^2 + 6\lambda + 1) + (49\lambda^2 + 28\lambda + 4) = 58\lambda^2 + 34\lambda + 5.$$

Thus $$f(\lambda) = \dfrac{(17\lambda + 5)^2} {58\lambda^2 + 34\lambda + 5}.$$ Take the derivative of $$f(\lambda)$$ (or use logarithmic differentiation):

Let $$g(\lambda) = \ln f(\lambda)$$.
$$g(\lambda) = 2\ln|17\lambda + 5| - \ln(58\lambda^2 + 34\lambda + 5).$$
Differentiate and set to zero:
$$g'(\lambda) = \frac{34}{17\lambda + 5} - \frac{116\lambda + 34}{58\lambda^2 + 34\lambda + 5} = 0.$$

Multiply through to clear denominators:
$$34(58\lambda^2 + 34\lambda + 5) - (17\lambda + 5)(116\lambda + 34) = 0.$$ Expanding both products gives
$$1972\lambda^2 + 1156\lambda + 170 - \bigl(1972\lambda^2 + 1158\lambda + 170\bigr) = 0,$$ which simplifies to $$-2\lambda = 0 \;\Rightarrow\; \lambda = 0.$$

For $$|\lambda|\to\infty$$, $$D(\lambda) \to \dfrac{17}{\sqrt{58}},$$ which is smaller than the value at $$\lambda = 0.$$ Hence $$\lambda = 0$$ indeed gives the farthest line.

Step 3 : Equation of the required line $$L$$
Put $$\lambda = 0$$ in the family:
$$(3\cdot0 + 1)x + (7\cdot0 + 2)y = 17\cdot0 + 5$$
$$\Rightarrow\; x + 2y = 5.$$

Step 4 : Distance of $$L$$ from the point $$(3,6)$$
For the line $$x + 2y - 5 = 0$$ and the point $$(x_1,y_1) = (3,6)$$,
the perpendicular distance is
$$d = \frac{|1\cdot3 + 2\cdot6 - 5|}{\sqrt{1^2 + 2^2}} = \frac{|3 + 12 - 5|}{\sqrt{5}} = \frac{10}{\sqrt{5}} = 2\sqrt{5}.$$

Step 5 : Required value of $$d^2$$
$$d^2 = (2\sqrt{5})^2 = 4 \times 5 = 20.$$

Therefore $$d^2 = 20,$$ corresponding to Option A.