In acidic medium, $$K_2Cr_2O_7$$ shows oxidising action as represented in the half reaction $$Cr_2O_7^{2-} + XH^+ + Ye^- \rightarrow 2A + ZH_2O$$. X, Y, Z and A are respectively:

Balance the half-reaction: $$Cr_2O_7^{2-} + XH^+ + Ye^- \rightarrow 2A + ZH_2O$$.

In acidic medium, $$Cr_2O_7^{2-}$$ is reduced to $$Cr^{3+}$$, each Cr going from oxidation state +6 to +3. Since there are two Cr atoms on the left, the product on the right must be 2$$Cr^{3+}$$, so A = $$Cr^{3+}$$.

To balance oxygen, note the 7 oxygen atoms in $$Cr_2O_7^{2-}$$; these require 7$$H_2O$$ on the right, giving Z = 7.

Those 7$$H_2O$$ molecules contribute 14 hydrogen atoms on the right, so we add 14$$H^+$$ on the left to balance hydrogen, yielding X = 14.

Next, balance charge by adding electrons. The left side has charge $$(-2) + 14(+1) + Y(-1) = 12 - Y$$, while the right side has charge $$2(+3) + 0 = +6$$. Setting these equal gives $$ 12 - Y = 6 \implies Y = 6 $$.

Thus the balanced equation is $$Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$$, and checking charges shows both sides sum to +6.

X = 14, Y = 6, Z = 7, A = $$Cr^{3+}$$.

The correct answer is Option D: 14, 6, 7 and $$Cr^{3+}$$.