Solution
\[
\begin{aligned}
& \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{a}} \times \hat{\mathrm{b}}=\left|\begin{array}{ccc}
\hat{\lambda} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\
2 & -1 & 1 \\
0 & \lambda & 2
\end{array}\right|=(-2-\lambda) \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+2 \lambda \hat{\mathrm{k}} \\
& |\overrightarrow{\mathrm{c}}|=\sqrt{53}=\sqrt{(-2-\lambda)^{2}+16+4 \lambda^{2}} \\
& \Rightarrow 53=4+\lambda^{2}+4 \lambda+16+4 \lambda^{2} \\
& \Rightarrow 5 \lambda^{2}+4 \lambda-33=0 \Rightarrow(5 \lambda-11)(\lambda+3)=0 \\
& \lambda=11 / 5(\text { reject }) \lambda=-3(\text { accept }) ; \lambda \in Z \\
& \overrightarrow{\mathrm{c}}=\hat{\mathrm{i}}-4 \hat{\mathrm{j}}-6 \hat{\mathrm{k}} \\
& \text { let } \overrightarrow{\mathrm{d}}=x \hat{\mathrm{i}}+\hat{\mathrm{j}}+z \hat{\mathrm{k}} \\
& \overrightarrow{\mathrm{~d}} \text { lies in yz plane, } x=0 \\
& \overrightarrow{\mathrm{~d}}=\hat{\mathrm{y}}+z \hat{\mathrm{k}} \\
& |\overrightarrow{\mathrm{~d}}|=2=\sqrt{\mathrm{y}^{2}+z^{2}} \Rightarrow \mathrm{y}^{2}+\mathrm{z}^{2}=4 \\
& \overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{~d}}=|\overrightarrow{\mathrm{c}} \| \overrightarrow{\mathrm{d}}| \cos \theta \\
& \Rightarrow(\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{~d}})^{2}=|\overrightarrow{\mathrm{c}}|^{2}|\overrightarrow{\mathrm{~d}}|^{2} \cos ^{2} \theta \leq|\overrightarrow{\mathrm{c}}|^{2}|\overrightarrow{\mathrm{~d}}|^{2} \\
& \Rightarrow(\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{~d}})^{2} \leq(53)(4)=212
\end{aligned}
\]
Clearly $\overrightarrow{\mathrm{c}}$ \& $\overrightarrow{\mathrm{d}}$ are not parallel
It means $(\overrightarrow{\mathrm{c}} . \overrightarrow{\mathrm{d}})^{2}<212$
\[
\begin{aligned}
& \overrightarrow{\mathrm{c}}=\hat{\mathrm{i}}-4 \hat{\mathrm{j}}-6 \hat{\mathrm{k}} ; \overrightarrow{\mathrm{d}}=\mathrm{y} \hat{\mathrm{j}}+\mathrm{zk} \\
& \overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{~d}}=-4 \mathrm{y}-6 \mathrm{z} \\
& \Rightarrow(4 \mathrm{y}+6 \mathrm{z})^{2} \leq\left(\sqrt{4^{2}+6^{2}} \sqrt{\mathrm{y}^{2}+\mathrm{z}^{2}}\right)^{2} \\
& \Rightarrow(\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{~d}})^{2} \leq(52)(4)=208 \text { Answer(3) }
\end{aligned}
\]
.
