Let $\vec a =\hat i +4\hat j +2\hat k, \vec b=3\hat i-2\hat j+7\hat k$ and $\vec c=2\hat i+\hat j+4\hat k$. A vector $\vec d$ which is perpendicular to both $\vec a$ and $\vec b$ and $\vec c.\vec d=14$, is:

$64\hat i-2\hat j-28\hat k$

The correct answer is Option (2) → $64\hat i-2\hat j-28\hat k$

Given:

$\vec{a} = \hat{i} + 4\hat{j} + 2\hat{k}$

$\vec{b} = 3\hat{i} - 2\hat{j} + 7\hat{k}$

$\vec{c} = 2\hat{i} + \hat{j} + 4\hat{k}$

Since $\vec{d}$ is perpendicular to both $\vec{a}$ and $\vec{b}$,

$\vec{d} \propto \vec{a} \times \vec{b}$

Compute $\vec{a} \times \vec{b}$:

$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 4 & 2 \\ 3 & -2 & 7 \end{vmatrix}$

$= \hat{i}(4 \cdot 7 - 2 \cdot (-2)) - \hat{j}(1 \cdot 7 - 2 \cdot 3) + \hat{k}(1 \cdot (-2) - 4 \cdot 3)$

$= \hat{i}(28 + 4) - \hat{j}(7 - 6) + \hat{k}(-2 - 12)$

$= 32\hat{i} - 1\hat{j} - 14\hat{k}$

So, $\vec{a} \times \vec{b} = 32\hat{i} - \hat{j} - 14\hat{k}$

Now, $\vec{c} \cdot \vec{d} = 14$.

Let $\vec{d} = \lambda(32\hat{i} - \hat{j} - 14\hat{k})$

Then, $\vec{c} \cdot \vec{d} = \lambda(2 \cdot 32 + 1 \cdot (-1) + 4 \cdot (-14))$

$\Rightarrow 14 = \lambda(64 - 1 - 56)$

$\Rightarrow 14 = \lambda(7)$

$\Rightarrow \lambda = 2$

Hence,

$\vec{d} = 2(32\hat{i} - \hat{j} - 14\hat{k})$

$\vec{d} = 64\hat{i} - 2\hat{j} - 28\hat{k}$

Therefore, $\vec{d} = 64\hat{i} - 2\hat{j} - 28\hat{k}$.