Let `veca=hati+2hatj+4hatk, vecb=hati+lambdahatj+4hatk` and `vecc=2hati+4hatj+(lambda^2-1)hatk` be coplanar vectors . Then the non-zero vector `veca xx vecc` is `-2lhati+5 hatj` . The value of l is ____.

Explanation

To solve the problem, we need to find the value of \( \lambda \) such that the vectors \( \vec{a} \), \( \vec{b} \), and \( \vec{c} \) are coplanar, and the cross product \( \vec{a} \times \vec{c} \) equals \( -2\hat{i} + 5\hat{j} \).

### Step 1: Write down the vectors
Given:
\[ \vec{a} = \hat{i} + 2\hat{j} + 4\hat{k} \]
\[
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