Problem 1161

Drop the altitude from $A$ to $BC$, and let $F$ be its foot. Furthermore, suppose that $BF = x$ (if $F$ lies on the extension of $BC$ beyond $B$, assign a negative sign to $x$) and $AF = y$. Then, by the Pythagorean theorem, $$AB^2 = BF^2 + AF^2 = x^2 + y^2,$$ $$AC^2 = CF^2 + AF^2 = (3-x)^2 + y^2,$$ $$AD^2 = DF^2 + AF^2 = (2-x)^2 + y^2.$$ It follows that $$\begin{aligned} AB^2 + 2AC^2 - 3AD^2 &= x^2 + y^2 + 2(3-x)^2 + 2y^2 - 3(2-x)^2 - 3y^2 \\ &= x^2 + y^2 + 18 - 12x + 2x^2 + 2y^2 - 12 + 12x - 3x^2 - 3y^2 \\ &= 6, \end{aligned}$$ regardless of the values of $x$ and $y$.