The locus of the foot of perpendicular drawn from the centre of the ellipse $$x^2 + 3y^2 = 6$$ on any tangent to it is:

We start with the given ellipse

$$x^{2}+3y^{2}=6.$$

It is convenient to write it in the standard form

$$\frac{x^{2}}{6}+\frac{y^{2}}{2}=1,$$

so that we can recognise the semi-axes $$a^{2}=6$$ and $$b^{2}=2.$$

For any point $$\,(x_{1},y_{1})$$ lying on the ellipse, i.e.

$$x_{1}^{2}+3y_{1}^{2}=6,$$

the equation of the tangent at that point is obtained by the tangent in the form $$T=0$$ formula:

$$xx_{1}+3yy_{1}=6.$$

The centre of the ellipse is the origin $$O(0,0).$$ We are required to find the locus of the foot of the perpendicular drawn from this origin to an arbitrary tangent. Let the foot of that perpendicular be $$P(h,k).$$

Because $$OP$$ is perpendicular to the tangent, the vector $$\overrightarrow{OP}=(h,k)$$ must be parallel to the normal vector of the tangent. For the line

$$xx_{1}+3yy_{1}=6,$$

the normal vector is clearly $$\bigl(x_{1},\,3y_{1}\bigr).$$ Hence there is some scalar $$t$$ such that

$$h=t\,x_{1},\qquad k=t\,(3y_{1}).$$

Point $$P(h,k)$$ itself lies on the tangent, so we substitute $$x=h,\;y=k$$ into the tangent’s equation:

$$x_{1}h+3y_{1}k=6.$$ Substituting $$h=t\,x_{1}$$ and $$k=t\,3y_{1}$$ gives

$$x_{1}(t\,x_{1})+3y_{1}(t\,3y_{1})=t\bigl(x_{1}^{2}+9y_{1}^{2}\bigr)=6.$$

Therefore

$$t=\frac{6}{x_{1}^{2}+9y_{1}^{2}}.$$

Using this value of $$t$$ we can express $$h$$ and $$k$$ completely in terms of $$x_{1},y_{1}:$$

$$h=\frac{6x_{1}}{x_{1}^{2}+9y_{1}^{2}},\qquad k=\frac{18y_{1}}{x_{1}^{2}+9y_{1}^{2}}.$$

Our goal is to eliminate $$x_{1},y_{1}$$ to obtain a relation between $$h$$ and $$k$$ only. To that end we let

$$D=x_{1}^{2}+9y_{1}^{2},$$

so that the above formulae become

$$h=\frac{6x_{1}}{D},\qquad k=\frac{18y_{1}}{D}.$$

From these we solve for $$x_{1}$$ and $$y_{1}$$:

$$x_{1}=\frac{hD}{6},\qquad y_{1}=\frac{kD}{18}.$$

Since $$\,(x_{1},y_{1})$$ lies on the ellipse, it satisfies

$$x_{1}^{2}+3y_{1}^{2}=6.$$

Substituting the expressions for $$x_{1}$$ and $$y_{1}$$ gives

$$\left(\frac{hD}{6}\right)^{2}+3\left(\frac{kD}{18}\right)^{2}=6.$$

Expanding, we obtain

$$\frac{h^{2}D^{2}}{36}+\frac{k^{2}D^{2}}{108}=6.$$

Factorising $$\dfrac{1}{108}$$ out of the brackets,

$$\frac{D^{2}}{108}\bigl(3h^{2}+k^{2}\bigr)=6,$$ so

$$D^{2}=\frac{648}{3h^{2}+k^{2}}.$$

On the other hand, by definition

$$D=x_{1}^{2}+9y_{1}^{2}=\frac{h^{2}D^{2}}{36}+\frac{k^{2}D^{2}}{36} = \frac{D^{2}(h^{2}+k^{2})}{36}.$$

Dividing both sides by $$D\ (\neq 0)$$ gives

$$1=\frac{D(h^{2}+k^{2})}{36},\qquad\text{so}\qquad D=\frac{36}{h^{2}+k^{2}}.$$

Squaring this last relation,

$$D^{2}=\frac{1296}{(h^{2}+k^{2})^{2}}.$$

We now have two separate expressions for $$D^{2}$$, so we equate them:

$$\frac{1296}{(h^{2}+k^{2})^{2}}=\frac{648}{3h^{2}+k^{2}}.$$

Cross-multiplying yields

$$1296\bigl(3h^{2}+k^{2}\bigr)=648\,(h^{2}+k^{2})^{2}.$$

Dividing by $$648$$ simplifies the equation to

$$2\bigl(3h^{2}+k^{2}\bigr)=(h^{2}+k^{2})^{2}.$$

That is,

$$(h^{2}+k^{2})^{2}=6h^{2}+2k^{2}.$$

Replacing $$h$$ by $$x$$ and $$k$$ by $$y$$ (for the locus in the usual $$xy$$-plane) we finally obtain

$$(x^{2}+y^{2})^{2}=6x^{2}+2y^{2}.$$

This exactly matches the equation given in Option A.

Hence, the correct answer is Option A.