Vector Algebra – Vector Products – JEE Main 24 Jan 2026 Shift 2

Solution:

$$2(\vec{a}\times\vec{c})+3(\vec{b}\times\vec{c})=\vec{0}$$
$$(2\vec{a} + 3\vec{b}) \times \vec{c} = \vec{0}$$

This implies $\vec{c}$ is parallel to $(2\vec{a} + 3\vec{b})$.
Let $\vec{v} = 2\vec{a} + 3\vec{b}$.
$$\vec{v} = 2(2\hat{i}-5\hat{j}+5\hat{k}) + 3(\hat{i}-\hat{j}+3\hat{k})$$
$$\vec{v} = (4\hat{i}-10\hat{j}+10\hat{k}) + (3\hat{i}-3\hat{j}+9\hat{k}) = 7\hat{i} – 13\hat{j} + 19\hat{k}$$

Let $\vec{c} = \lambda \vec{v} = \lambda(7\hat{i} – 13\hat{j} + 19\hat{k})$.

Given $(\vec{a}-\vec{b}) \cdot \vec{c} = -97$:
$$\vec{a} – \vec{b} = (2-1)\hat{i} + (-5+1)\hat{j} + (5-3)\hat{k} = \hat{i} – 4\hat{j} + 2\hat{k}$$
$$(\hat{i} – 4\hat{j} + 2\hat{k}) \cdot \lambda(7\hat{i} – 13\hat{j} + 19\hat{k}) = -97$$
$$\lambda [1(7) + (-4)(-13) + 2(19)] = -97$$
$$\lambda [7 + 52 + 38] = -97 \Rightarrow 97\lambda = -97 \Rightarrow \lambda = -1$$

So $\vec{c} = -1(7\hat{i} – 13\hat{j} + 19\hat{k}) = -7\hat{i} + 13\hat{j} – 19\hat{k}$.

We need to find $|\vec{c} \times \hat{k}|^2$.
$$\vec{c} \times \hat{k} = (-7\hat{i} + 13\hat{j} – 19\hat{k}) \times \hat{k} = -7(\hat{i}\times\hat{k}) + 13(\hat{j}\times\hat{k}) – 19(\hat{k}\times\hat{k})$$
$$\vec{c} \times \hat{k} = -7(-\hat{j}) + 13(\hat{i}) – 0 = 13\hat{i} + 7\hat{j}$$

$$|\vec{c} \times \hat{k}|^2 = (13)^2 + (7)^2 = 169 + 49 = 218$$

Ans. (3)