Q.Three vectors $\vec{A}$, $\vec{B}$ and $\vec{C}$ add up to zero. Find which is false.

Concept: Vector triple products and the constraint $\vec{A} + \vec{B} + \vec{C} = \vec{0}$.

Since $\vec{A} + \vec{B} + \vec{C} = \vec{0}$, we have $\vec{C} = -(\vec{A} + \vec{B})$. The three vectors form a closed triangle (or are collinear).

Check each option:

(A) $(\vec{A} \times \vec{B}) \times \vec{C}$ lies in the plane of $\vec{A}$ and $\vec{B}$ by the BAC-CAB rule. It is zero only when $\vec{A} \times \vec{B} = \vec{0}$ (i.e., $\vec{A} \parallel \vec{B}$) or when $\vec{C}$ is parallel to $\vec{A} \times \vec{B}$ (i.e., $\vec{C}$ is perpendicular to the plane of $\vec{A}$, $\vec{B}$). The condition "$\vec{B}$, $\vec{C}$ parallel" is neither necessary nor sufficient. False.

(B) $(\vec{A} \times \vec{B}) \cdot \vec{C}$ is the scalar triple product, which equals zero if and only if the three vectors are coplanar. Since they sum to zero, they always lie in a plane (the triangle they form). Thus $(\vec{A} \times \vec{B}) \cdot \vec{C} = 0$ always, regardless of whether $\vec{B} \parallel \vec{C}$. False.

(C) As noted, $(\vec{A} \times \vec{B}) \times \vec{C}$ lies in the plane of $\vec{A}$ and $\vec{B}$, which is the same plane containing all three vectors. True. …

The key consequence of $\vec{A}+\vec{B}+\vec{C}=\vec{0}$

Three vectors summing to zero form a closed triangle, so they lie in one plane - they are coplanar. Also $\vec{C} = -(\vec{A}+\vec{B})$.

Statement (B) - the false one

$(\vec{A}\times\vec{B})\cdot\vec{C}$ is the scalar triple product, the volume of the parallelepiped on $\vec{A},\vec{B},\vec{C}$. For coplanar vectors this volume is zero. Directly:

because $\vec{A}\times\vec{B}$ is perpendicular to both $\vec{A}$ and $\vec{B}$. It is always zero, not "non-zero unless $\vec{B},\vec{C}$ are parallel." Hence (B) is false.

Why the others are true

(A) By the triple-product identity $(\vec{A}\times\vec{B})\times\vec{C} = \vec{B}(\vec{A}\cdot\vec{C}) - \vec{A}(\vec{B}\cdot\vec{C})$. Its magnitude equals $|\vec{A}\times\vec{B}|\,|\vec{C}|$ (as $\vec{A}\times\vec{B}\perp\vec{C}$), which vanishes only when $\vec{A}\parallel\vec{B}$; but with $\vec{A}+\vec{B}+\vec{C}=\vec{0}$ that forces all three collinear, i.e. $\vec{B}\parallel\vec{C}$ as well. So the statement holds - (A) is true. …

Concept: Settle the Triple Products with One Concrete Coplanar Triangle

Method: Explicit Coordinate Construction (Pick Real Numbers for $\vec A,\vec B,\vec C$), Not Symbolic BAC-CAB Expansion

Rather than expanding $(\vec A\times\vec B)\times\vec C$ symbolically via the BAC-CAB identity, this method picks one concrete, easy-to-visualize triple of vectors satisfying $\vec A+\vec B+\vec C=\vec 0$ and computes every quantity in all four options directly as numbers -- turning an abstract vector-identity question into ordinary arithmetic.

Step 1 -- Build an explicit example

Since any three vectors summing to zero form a closed triangle, take a simple right triangle in the $xy$-plane:

Check: $\vec A+\vec B+\vec C = (3-3)\hat i + (3-3)\hat j = \vec 0$. Good -- this is a legitimate instance of the problem's hypothesis, and the three vectors are visibly coplanar (all lie in the $xy$-plane, with $\hat k$-component zero).

Step 2 -- Compute $(\vec A\times\vec B)$ once, reuse it in every option

Step 3 -- Test option (B): $(\vec A\times\vec B)\cdot\vec C$

The claim in (B) is that this is non-zero unless $\vec B,\vec C$ are parallel -- but here $\vec B=3\hat j$ and $\vec C=-3\hat i-3\hat j$ are clearly not parallel (different directions), and yet the scalar triple product is exactly $0$. This single computed number directly contradicts (B). (B) is false -- confirmed by direct arithmetic, no identity needed.

Step 4 -- Test option (A): $(\vec A\times\vec B)\times\vec C$

This is manifestly non-zero, and $\vec B=3\hat j$, $\vec C=-3\hat i-3\hat j$ are not parallel in this example -- consistent with (A)'s claim.

Step 5 -- Test option (C): does $(\vec A\times\vec B)\times\vec C = 27\hat i - 27\hat j$ lie in the plane of $\vec A,\vec B,\vec C$?

All three original vectors lie in the $xy$-plane (zero $\hat k$-component); the computed result $27\hat i-27\hat j$ also has zero $\hat k$-component. It lies in the same plane, consistent with (C)'s claim.

Step 6 -- Test option (D) …