If $\vec a = 2\hat i + m\hat j-n\hat k$ and $\vec b = l\hat i-3\hat j+4\hat k$ such that $\vec a = 2\vec b$ then the value of $14l +m+n$ is:

0

The correct answer is Option (1) → 0

Given:

$\vec{a} = 2\hat{i} + m\hat{j} - n\hat{k}$

$\vec{b} = l\hat{i} - 3\hat{j} + 4\hat{k}$

and $\vec{a} = 2\vec{b}$

Comparing components:

$2 = 2l \Rightarrow l = 1$

$m = 2(-3) \Rightarrow m = -6$

$-n = 2(4) \Rightarrow n = -8$

Compute:

$14l + m + n = 14(1) + (-6) + (-8) = 14 - 6 - 8 = 0$

Therefore, $14l + m + n = 0$.