If $\vec a = 3\hat i-6\hat j +\hat k$ and $\vec b = 2\hat i-4\hat j+λ\hat k$ are such that $\vec a||\vec b$, then $3λ + 2 =$

4

The correct answer is Option (3) → 4

Given vectors:

\(\vec{a} = 3\hat{i} - 6\hat{j} + \hat{k}\)

\(\vec{b} = 2\hat{i} - 4\hat{j} + \lambda \hat{k}\)

Since \(\vec{a} \parallel \vec{b}\), there exists a scalar \(k\) such that:

\(\vec{b} = k \vec{a}\)

Equate components:

\[ \begin{cases} 2 = 3k \\ -4 = -6k \\ \lambda = k \end{cases} \]

From first equation:

\(k = \frac{2}{3}\)

From second equation:

\(-4 = -6k \Rightarrow k = \frac{4}{6} = \frac{2}{3}\)

Both equal \(k = \frac{2}{3}\), consistent.

From third equation:

\(\lambda = k = \frac{2}{3}\)

Calculate \(3\lambda + 2\):

\[ 3 \times \frac{2}{3} + 2 = 2 + 2 = 4 \]