JEE Main 2026 — 21 January, Morning Shift — Question 5

Given $\vec{a} = -\hat{i} + 2\hat{j} + 2\hat{k}$, $\vec{b} = 8\hat{i} + 7\hat{j} - 3\hat{k}$. Let $\vec{c} = c_1\hat{i} + c_2\hat{j} + c_3\hat{k}$. $\vec{a} \times \vec{c} = \vec{b}$ gives: $(2c_3 - 2c_2)\hat{i} + (c_3 + 2c_1)\hat{j} - (c_2 + 2c_1)\hat{k} = 8\hat{i} + 7\hat{j} - 3\hat{k}$. Thus, $2c_3 - 2c_2 = 8$, $c_3 + 2c_1 = 7$, $c_2 + 2c_1 = 3$. Also, $\vec{c} \cdot (\hat{i} + \hat{j} + \hat{k}) = 4$ gives $c_1 + c_2 + c_3 = 4$. Solving: $c_1 = 2$, $c_2 = -1$, $c_3 = 3$. Then $\vec{a} + \vec{c} = \hat{i} + \hat{j} + 5\hat{k}$. $|\vec{a} + \vec{c}|^2 = 1^2 + 1^2 + 5^2 = 27$. Thus, the answer is $27.$