\(\begin{aligned}
\text {Let } \mathbf{a} & =\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}} \\
\mathbf{b} & =2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}-5 \hat{\mathbf{k}} \\
\mathbf{c} & =\lambda \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}} \\
\mathbf{b}+\mathbf{c} & =(2+\lambda) \hat{\mathbf{i}}+6 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}
\end{aligned}\)
\(\hat{\mathbf{d}}=\) unit vector along \((\mathbf{b}+\mathbf{c})=\frac{\mathbf{b}+\mathbf{c}}{|\mathbf{b}+\mathbf{c}|}\)
\(\begin{aligned}= & \frac{(2+\lambda) \hat{\mathbf{i}}+6 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}}{\sqrt{(2+\lambda)^2+36+4}} \\ \mathbf{d} & =\frac{(2+\lambda) \hat{\mathbf{i}}+6 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}}{\sqrt{\lambda^2+4 \lambda+44}}\end{aligned}\)
\(\begin{aligned} & \hat{\mathbf{a}} \times \hat{\mathbf{d}} \\ & =\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & 1 & 1 \\ \frac{(2+\lambda)}{\sqrt{\lambda^2+4 \lambda+44}} & \frac{6}{\sqrt{\lambda^2+4 \lambda+44}} & \frac{-2}{\sqrt{\lambda^2+4 \lambda+44}}\end{array}\right|\end{aligned}\)
\(\begin{aligned}
& =\frac{1}{\sqrt{\lambda^2+4 \lambda+44}}\left|\begin{array}{ccc}
\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\
1 & 1 & 1 \\
2+\lambda & 6 & -2
\end{array}\right| \\
& =\frac{1}{\sqrt{\lambda^2+4 \lambda+44}}|\hat{\mathbf{i}}(-8)-\hat{\mathbf{j}}(-2-2-\lambda)+\hat{\mathbf{k}}(6-2-\lambda)| \\
& \hat{\mathbf{a}} \times \hat{\mathbf{d}}=\frac{1}{\sqrt{\lambda^2+4 \lambda+44}}|-8 \hat{\mathbf{i}}+(4+\lambda) \mathbf{j}+(4-\lambda) \hat{\mathbf{k}}| \\
& |\hat{\mathbf{a}} \times \hat{\mathbf{d}}|=\frac{1}{\sqrt{\lambda^2+4 \lambda+44}} \sqrt{64+(4+\lambda)^2+(4-\lambda)^2} \\
& \sqrt{2}=\frac{\sqrt{64+16+\lambda^2+16+\lambda^2}}{\sqrt{\lambda^2+4 \lambda+44}}
\end{aligned}\)
Squaring on both sides,
\(2=\frac{2 \lambda^2+96}{\lambda^2+4 \lambda+44}\)
\(\begin{aligned}
2 \lambda^2+8 \lambda+88 & =2 \lambda^2+96 \\
8 \lambda & =8 \\
\lambda & =1
\end{aligned}\)
Hence, option (b) is correct.