Vector Algebra Part 3

Concept:Use vector cross product to find $\vec{c}$, then apply Cauchy-Schwarz inequality to maximize $(\vec{c}\cdot\vec{d})^2$ under the constraint that $\vec{d}$ lies in the $yz$-plane with $|\vec{d}|=2$.Explanation:Compute $\vec{c} = \vec{a}\times\vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 1 \\ 0 & \lambda & 2 \end{vmatrix} = (-2-\lambda)\hat{i} -4\hat{j} + 2\lambda\hat{k}$.Given $|\vec{c}|^2 = 53$: $(-2-\lambda)^2+16+4\lambda^2=53$ simplifies to $5\lambda^2+4\lambda-33=0$.Since $\lambda \in \mathbb{Z}$, solving gives $\lambda = -3$.Thus $\vec{c} = (1,\,-4,\,-6)$.Let $\vec{d}=(0,y,z)$ (in $yz$-plane) with $y^2+z^2=4$.Then $\vec{c}\cdot\vec{d} = -4y -6z$.Maximize $(\vec{c}\cdot\vec{d})^2$ by Cauchy-Schwarz: $(-4y-6z)^2 \le [(-4)^2+(-6)^2](y^2+z^2) = (16+36)\cdot4 = 208$.Equality is achievable when $(y,z)$ is proportional to $(-4,-6)$, e.g., $y=-\frac{4}{\sqrt{13}},\,z=-\frac{6}{\sqrt{13}}$.Hence the maximum possible value is $208$.Answer:$208$ (Option D)