Vector Algebra – Vector Product – JEE Main 23 Jan 2026 Shift 2

Solution:

First, we find $\vec{v} = \vec{a} \times \vec{b}$:
$$ \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 3 \\ 2 & 1 & -1 \end{vmatrix} = \hat{i}(2-3) – \hat{j}(-1-6) + \hat{k}(1+4) $$

$$ \vec{v} = -\hat{i} + 7\hat{j} + 5\hat{k} $$

Given $\vec{v} \cdot \vec{c} = 11$:
$$ (-\hat{i} + 7\hat{j} + 5\hat{k}) \cdot (\lambda\hat{i} + \hat{j} + \hat{k}) = 11 $$

$$ -\lambda + 7 + 5 = 11 \Rightarrow -\lambda + 12 = 11 \Rightarrow \lambda = 1 $$

So, $\vec{c} = \hat{i} + \hat{j} + \hat{k}$.
Now, projection of $\vec{b}$ on $\vec{c}$ is given by $p$:
$$ p = \frac{|\vec{b} \cdot \vec{c}|}{|\vec{c}|} $$

$$ \vec{b} \cdot \vec{c} = (2)(1) + (1)(1) + (-1)(1) = 2 $$
$$ |\vec{c}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3} $$

$$ \therefore p = \frac{2}{\sqrt{3}} $$

We need $9p^2$:
$$ 9p^2 = 9 \left( \frac{4}{3} \right) = 12 $$

Ans. (4)