No. 

This is an example with irreducible (and nonsingular) $A,B,C$: Consider $\mathbb{A}^4$ with coordinates $(w,x, y, z)$. Let $B$ be the $(x,y)$-plane (i.e. the set $w = z = 0$), $C$ be the hypersurface $z = xy$, and $A$ be the $(y,z)$-plane. Then $B \cap C$ is the union of the "$x$-axis" and "$y$-axis". Let $V_1$ be "$x$-axis" and $V$ be the origin, which is the only irreducible component of $A \cap V_1$. Since $A \cap B$ is the "$y$-axis", which is irreducible, the only possibility for $W_1$ is $y$-axis. But $W_1 \cap C = W_1 \neq V$. 

Note that each of $A, B, C$ is isomorphic to $\mathbb{A}^k$ (for an appropriate $k$). 

<b> Original Example:</b> Consider $\mathbb{A}^4$ with coordinates $(w,x, y, z)$. Let $B$ be the $(x,y)$-plane (i.e. the set $w = z = 0$), $C$ be the union of the $(w,x)$-plane and the $(w,y)$-plane and $A$ be the $(y,z)$-plane. Then $B \cap C$ is the union of the "$x$-axis" and "$y$-axis", whereas $A \cap B = A \cap C = A \cap B \cap C$ is the "$y$-axis", so that the property fails with $V_1$ equal to the "$x$-axis".