Let $$\overrightarrow{a}= 2\widehat{i}-\widehat{j}+\widehat{k}$$ and $$\overrightarrow{b}= \lambda \widehat{j}+2\widehat{k}, \lambda\in Z$$ be two vectors. Let $$\overrightarrow{c}= \overrightarrow{a} \times \overrightarrow{b} \text{and } \overrightarrow{d}$$ be a vector of magnitude 2 in yz-plane. If $$|\overrightarrow{c}|=\sqrt{53}$$, then the maximum possible value of $$\left(\overrightarrow{c}\cdot\overrightarrow{d}\right)^{2}$$ is equal to :

We need to find the maximum value of $$(\vec{c} \cdot \vec{d})^2$$ given the constraints.

First, we compute $$\vec{c} = \vec{a} \times \vec{b}$$. We have $$\vec{a} = 2\hat{i} - \hat{j} + \hat{k} = (2, -1, 1)$$ and $$\vec{b} = \lambda\hat{j} + 2\hat{k} = (0, \lambda, 2)$$. The cross product can be evaluated as:

$$\vec{c} = \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 1 \\ 0 & \lambda & 2 \end{vmatrix}$$

$$= \hat{i}[(-1)(2) - (1)(\lambda)] - \hat{j}[(2)(2) - (1)(0)] + \hat{k}[(2)(\lambda) - (-1)(0)]$$

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

$$= (-2-\lambda, -4, 2\lambda)$$

Next, using $$|\vec{c}| = \sqrt{53}$$ we determine $$\lambda$$ by setting $$|\vec{c}|^2 = (-2-\lambda)^2 + (-4)^2 + (2\lambda)^2 = 53$$, which gives:

$$ (\lambda+2)^2 + 16 + 4\lambda^2 = 53$$

$$ \lambda^2 + 4\lambda + 4 + 16 + 4\lambda^2 = 53$$

$$ 5\lambda^2 + 4\lambda + 20 = 53$$

$$ 5\lambda^2 + 4\lambda - 33 = 0$$

By the quadratic formula, $$\lambda = \frac{-4 \pm \sqrt{16 + 660}}{10} = \frac{-4 \pm \sqrt{676}}{10} = \frac{-4 \pm 26}{10}$$, so $$\lambda = \frac{22}{10} = 2.2$$ or $$\lambda = \frac{-30}{10} = -3$$. Since $$\lambda \in \mathbb{Z}$$, it follows that $$\lambda = -3$$.

Substituting $$\lambda = -3$$ into the expression for $$\vec{c}$$ yields $$\vec{c} = (-2-(-3), -4, 2(-3)) = (1, -4, -6)$$.

Finally, to maximize $$(\vec{c} \cdot \vec{d})^2$$, note that $$\vec{d}$$ lies in the yz-plane with $$|\vec{d}| = 2$$, so we write $$\vec{d} = (0, d_2, d_3)$$ subject to $$d_2^2 + d_3^2 = 4$$. Then

$$\vec{c} \cdot \vec{d} = (1)(0) + (-4)(d_2) + (-6)(d_3) = -4d_2 - 6d_3$$

and we seek the maximum of $$(-4d_2 - 6d_3)^2$$ under the constraint $$d_2^2 + d_3^2 = 4$$. By the Cauchy-Schwarz inequality:

$$(-4d_2 - 6d_3)^2 \leq ((-4)^2 + (-6)^2)(d_2^2 + d_3^2) = (16 + 36)(4) = 52 \times 4 = 208$$

Equality occurs when $$\frac{d_2}{-4} = \frac{d_3}{-6}$$, i.e.\ when $$\vec{d}$$ is parallel to the projection of $$\vec{c}$$ onto the yz-plane, which is possible since $$\vec{d}$$ lies in that plane. Hence the maximum value of $$(\vec{c} \cdot \vec{d})^2$$ is 208.

The correct answer is Option 1: 208.