If $$\vec{u} = \hat{j} + 4\hat{k}$$, $$\vec{v} = \hat{i} + 3\hat{k}$$ and $$\vec{w} = \cos\theta\hat{i} + \sin\theta\hat{j}$$ are vectors in 3-dimensional space, then the maximum possible value of $$|\vec{u} \times \vec{v} \cdot \vec{w}|$$ is

Write the three vectors in component form:
$$\vec{u} = 0\,\hat{i} + 1\,\hat{j} + 4\,\hat{k},\qquad \vec{v} = 1\,\hat{i} + 0\,\hat{j} + 3\,\hat{k},\qquad \vec{w} = \cos\theta\,\hat{i} + \sin\theta\,\hat{j} + 0\,\hat{k}.$$ (The $$\hat{k}$$-component of $$\vec{w}$$ is $$0$$ because it lies in the $$xy$$-plane.)

Step 1: Find the cross product $$\vec{u}\times\vec{v}.$$ $$ \vec{u}\times\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 0 & 1 & 4\\ 1 & 0 & 3 \end{vmatrix} = \hat{i}(1\cdot 3 - 4\cdot 0) - \hat{j}(0\cdot 3 - 4\cdot 1) + \hat{k}(0\cdot 0 - 1\cdot 1) = 3\,\hat{i} + 4\,\hat{j} - 1\,\hat{k}. $$

Step 2: Form the scalar triple product with $$\vec{w}.$$ $$ (\vec{u}\times\vec{v})\cdot\vec{w} = (3\,\hat{i} + 4\,\hat{j} - 1\,\hat{k})\cdot(\cos\theta\,\hat{i} + \sin\theta\,\hat{j} + 0\,\hat{k}) = 3\cos\theta + 4\sin\theta. $$

Step 3: Maximise the absolute value of $$3\cos\theta + 4\sin\theta.$$ For any real numbers $$a$$ and $$b$$, the expression $$a\cos\theta + b\sin\theta$$ attains values between $$-\sqrt{a^2+b^2}$$ and $$+\sqrt{a^2+b^2}.$$ Here $$a = 3$$ and $$b = 4,$$ so $$ \sqrt{a^2 + b^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5. $$ Hence $$ \max_\theta \bigl|3\cos\theta + 4\sin\theta\bigr| = 5. $$ Equality occurs when $$\tan\theta = \tfrac{4}{3}.$

Therefore the maximum possible value of $$|$$\vec{u}$$$$\times$$$$\vec{v}$$$$\cdot$$$$\vec{w}$$|$$ is $$5$$.

Option B which is: $$5$$