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(B) Let $\vec{a} = \hat{i}+\hat{j}+\hat{k}$,$\vec{b} = 2 \hat{i}+4 \hat{j}-5 \hat{k}$,and $\vec{c} = \lambda \hat{i}+2 \hat{j}+3 \hat{k}$.Then $\vec{b

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$1$

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(B) Let $\vec{a} = \hat{i}+\hat{j}+\hat{k}$,$\vec{b} = 2 \hat{i}+4 \hat{j}-5 \hat{k}$,and $\vec{c} = \lambda \hat{i}+2 \hat{j}+3 \hat{k}$.Then $\vec{b}+\vec{c} = (2+\lambda) \hat{i}+6 \hat{j}-2 \hat{k}$.Let $\hat{u}$ be the unit vector along $(\vec{b}+\vec{c})$,so $\hat{u} = \frac{\vec{b}+\vec{c}}{|\vec{b}+\vec{c}|} = \frac{(2+\lambda) \hat{i}+6 \hat{j}-2 \hat{k}}{\sqrt{(2+\lambda)^2+36+4}} = \frac{(2+\lambda) \hat{i}+6 \hat{j}-2 \hat{k}}{\sqrt{\lambda^2+4 \lambda+44}}$.The magnitude of the vector product is $|\vec{a} 	imes \hat{u}| = \sqrt{2}$.$\vec{a} 	imes \hat{u} = \frac{1}{\sqrt{\lambda^2+4 \lambda+44}} \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & 1 \ 2+\lambda & 6 & -2 \end{vmatrix} = \frac{1}{\sqrt{\lambda^2+4 \lambda+44}} [\hat{i}(-2-6) - \hat{j}(-2-(2+\lambda)) + \hat{k}(6-(2+\lambda))]$.$= \frac{1}{\sqrt{\lambda^2+4 \lambda+44}} [-8 \hat{i} + (4+\lambda) \hat{j} + (4-\lambda) \hat{k}]$.Taking the magnitude: $|\vec{a} 	imes \hat{u}| = \frac{\sqrt{(-8)^2 + (4+\lambda)^2 + (4-\lambda)^2}}{\sqrt{\lambda^2+4 \lambda+44}} = \sqrt{2}$.$\frac{\sqrt{64 + 16 + 8\lambda + \lambda^2 + 16 - 8\lambda + \lambda^2}}{\sqrt{\lambda^2+4 \lambda+44}} = \sqrt{2} \implies \frac{\sqrt{2\lambda^2 + 96}}{\sqrt{\lambda^2+4 \lambda+44}} = \sqrt{2}$.Squaring both sides: $\frac{2\lambda^2 + 96}{\lambda^2+4 \lambda+44} = 2 \implies 2\lambda^2 + 96 = 2\lambda^2 + 8\lambda + 88$.$8\lambda = 8 \implies \lambda = 1$. Thus,option $(B)$ is correct.

If the magnitude of the vector product of the vector $\hat{i}+\hat{j}+\hat{k}$ with a unit vector along the sum of the vectors $2 \hat{i}+4 \hat{j}-5 \hat{k}$ and $\lambda \hat{i}+2 \hat{j}+3 \hat{k}$ is equal to $\sqrt{2}$,then the value of ' $\lambda$ ' is

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