Let $$\vec{a} = \hat{i} + 2\hat{j} + 4\hat{k}$$, $$\vec{b} = \hat{i} + \lambda\hat{j} + 4\hat{k}$$ and $$\vec{c} = 2\hat{i} + 4\hat{j} + (\lambda^2 - 1)\hat{k}$$ be coplanar vectors. Then the non-zero vector $$\vec{a} \times \vec{c}$$ is:

We have the three vectors

$$\vec a = \hat i + 2\hat j + 4\hat k,\qquad \vec b = \hat i + \lambda\hat j + 4\hat k,\qquad \vec c = 2\hat i + 4\hat j + (\lambda^2-1)\hat k.$$

Since the vectors are given to be coplanar, their scalar triple product must be zero. The scalar triple product formula is

$$\vec a\cdot(\vec b\times\vec c)=0.$$

First we compute the cross product $$\vec b\times\vec c.$$ Using the determinant form of the cross product,

$$ \vec b\times\vec c=\begin{vmatrix} \hat i & \hat j & \hat k\\ 1 & \lambda & 4\\ 2 & 4 & \lambda^2-1 \end{vmatrix}. $$

Expanding the determinant along the first row:

$$ \vec b\times\vec c = \hat i\bigl[\lambda(\lambda^2-1)-4\cdot4\bigr] -\hat j\bigl[1(\lambda^2-1)-4\cdot2\bigr] +\hat k\bigl[1\cdot4-\lambda\cdot2\bigr]. $$

Now we simplify each component.

$$\lambda(\lambda^2-1)-16=\lambda^3-\lambda-16,$$ $$1(\lambda^2-1)-8=\lambda^2-1-8=\lambda^2-9,$$ $$4-2\lambda.$$

Thus

$$\vec b\times\vec c=(\lambda^3-\lambda-16)\hat i-(\lambda^2-9)\hat j+(4-2\lambda)\hat k.$$

Next we take the dot product of this with $$\vec a=1\hat i+2\hat j+4\hat k.$$

$$ \vec a\cdot(\vec b\times\vec c) =1\cdot(\lambda^3-\lambda-16) +2\cdot\bigl[-(\lambda^2-9)\bigr] +4\cdot(4-2\lambda). $$

Carrying out the multiplication term by term:

$$1(\lambda^3-\lambda-16)=\lambda^3-\lambda-16,$$ $$2\bigl[-(\lambda^2-9)\bigr]=-2\lambda^2+18,$$ $$4(4-2\lambda)=16-8\lambda.$$

Adding all these contributions gives

$$\lambda^3-\lambda-16-2\lambda^2+18+16-8\lambda =\lambda^3-2\lambda^2-9\lambda+18.$$

For coplanarity this expression must be zero, so

$$\lambda^3-2\lambda^2-9\lambda+18=0.$$

We factor the cubic. Trying simple factors, $$\lambda=3$$ satisfies the equation, so we divide by $$(\lambda-3)$$:

$$\lambda^3-2\lambda^2-9\lambda+18=(\lambda-3)(\lambda^2+\lambda-6).$$

Further factoring $$\lambda^2+\lambda-6=(\lambda+3)(\lambda-2).$$

Hence

$$\lambda^3-2\lambda^2-9\lambda+18=(\lambda-3)(\lambda+3)(\lambda-2)=0,$$

giving three possible values

$$\lambda=3,\qquad\lambda=-3,\qquad\lambda=2.$$

We now turn to the required cross product $$\vec a\times\vec c.$$ Again we use the determinant:

$$ \vec a\times\vec c=\begin{vmatrix} \hat i & \hat j & \hat k\\ 1 & 2 & 4\\ 2 & 4 & \lambda^2-1 \end{vmatrix}. $$

Expanding along the first row:

$$ \vec a\times\vec c = \hat i\bigl[2(\lambda^2-1)-4\cdot4\bigr] -\hat j\bigl[1(\lambda^2-1)-4\cdot2\bigr] +\hat k\bigl[1\cdot4-2\cdot2\bigr]. $$

We simplify each bracket:

$$2(\lambda^2-1)-16=2\lambda^2-2-16=2\lambda^2-18,$$ $$1(\lambda^2-1)-8=\lambda^2-9,$$ $$4-4=0.$$

Therefore

$$\vec a\times\vec c=(2\lambda^2-18)\hat i-(\lambda^2-9)\hat j+0\hat k.$$

Factoring out the common term $$(\lambda^2-9)$$:

$$\vec a\times\vec c=(\lambda^2-9)\bigl(2\hat i-\hat j\bigr).$$

This result shows that if $$\lambda^2-9=0$$ (i.e., $$\lambda=\pm3$$) the cross product vanishes, but the question asks for a non-zero vector. Hence we must choose the remaining admissible value $$\lambda=2.$$

Substituting $$\lambda=2$$ gives

$$\lambda^2-9=4-9=-5,$$

and hence

$$\vec a\times\vec c=-5\,(2\hat i-\hat j)=-10\hat i+5\hat j.$$

This matches the vector in Option D.

Hence, the correct answer is Option D.