If the lines vec{r} = 2hat{i} + hat{j} + hat{ | vedclass

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(D) The equations of the lines are given by:$L_1: \vec{r} = (2 + \lambda)\hat{i} + (1 - 2\lambda)\hat{j} + \hat{k}$$L_2: \vec{r} = \hat{i} + (1 + \mu)

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(D) The equations of the lines are given by:$L_1: \vec{r} = (2 + \lambda)\hat{i} + (1 - 2\lambda)\hat{j} + \hat{k}$$L_2: \vec{r} = \hat{i} + (1 + \mu)\hat{j} + (-3 + 2\mu)\hat{k}$If the lines intersect,there must exist values of $\lambda$ and $\mu$ such that the coordinates are equal:$2 + \lambda = 1 \Rightarrow \lambda = -1$$1 - 2\lambda = 1 + \mu$$1 = -3 + 2\mu \Rightarrow 2\mu = 4 \Rightarrow \mu = 2$Substituting $\lambda = -1$ into the second equation: $1 - 2(-1) = 1 + 2 = 3$,and $1 + \mu = 1 + 2 = 3$. Since both sides are equal,the lines intersect.Thus,$\lambda + \mu = -1 + 2 = 1$.

If the lines $\vec{r} = 2\hat{i} + \hat{j} + \hat{k} + \lambda(\hat{i} - 2\hat{j})$ and $\vec{r} = \hat{i} + \hat{j} - 3\hat{k} + \mu(\hat{j} + 2\hat{k})$ intersect each other,then $(\lambda + \mu)$ is equal to

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