A variable plane passes through a fixed point (3, 2, 1) and meets x, y and z axes at A, B and C respectively. A plane is drawn parallel to yz-plane through A, a second plane is drawn parallel zx plane through B and a third plane is drawn parallel to xy-plane through C. Then the locus of the point of intersection of these three planes, is:

Let us denote the required variable plane by its intercept form. A general plane that cuts the x-, y- and z-axes at the points $$A(a,0,0),\;B(0,b,0),\;C(0,0,c)$$ is written as

$$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1.$$

This variable plane is given to pass through the fixed point $$P_{0}(3,\,2,\,1).$$ Substituting $$(x,y,z)=(3,2,1)$$ in the equation of the plane, we obtain the relation among the intercepts:

$$\frac{3}{a}+\frac{2}{b}+\frac{1}{c}=1.$$

Now we construct three new planes:

1. A plane through $$A(a,0,0)$$ parallel to the $$yz$$-plane. The $$yz$$-plane itself is given by $$x=0.$$ Being parallel, the required plane keeps the same normal and therefore has the equation

$$x=a.$$

2. A plane through $$B(0,b,0)$$ parallel to the $$zx$$-plane. The $$zx$$-plane is $$y=0,$$ so the required parallel plane is

$$y=b.$$

3. A plane through $$C(0,0,c)$$ parallel to the $$xy$$-plane. The $$xy$$-plane is $$z=0,$$ hence the parallel plane is

$$z=c.$$

The sought point $$Q$$ is the common intersection of these three mutually perpendicular planes. Solving their simultaneous equations gives directly

$$Q\equiv(a,\,b,\,c).$$

Thus the coordinates of the moving point are $$x=a,\;y=b,\;z=c.$$ We already have the single relation connecting $$a,\,b,\,c$$:

$$\frac{3}{a}+\frac{2}{b}+\frac{1}{c}=1.$$

Replacing $$a,\,b,\,c$$ by $$x,\,y,\,z$$ respectively, the locus of $$Q$$ in Cartesian form becomes

$$\frac{3}{x}+\frac{2}{y}+\frac{1}{z}=1.$$

Comparing with the given choices, we recognise this as Option C.

Hence, the correct answer is Option C.