Let overrightarrow{a} = hat{i} + 2hat{j} + ha | vedclass

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(A) The line $L_1$ is given by $\overrightarrow{r} = (-\hat{i} + 2\hat{j} + \hat{k}) + \lambda(\hat{i} + 2\hat{j} + \hat{k}) = (\lambda - 1)\hat{i} +

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$(8, 26, 12)$

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(A) The line $L_1$ is given by $\overrightarrow{r} = (-\hat{i} + 2\hat{j} + \hat{k}) + \lambda(\hat{i} + 2\hat{j} + \hat{k}) = (\lambda - 1)\hat{i} + (2\lambda + 2)\hat{j} + (\lambda + 1)\hat{k}$.The line $L_2$ is given by $\overrightarrow{r} = (\hat{j} + \hat{k}) + \mu(2\hat{i} + 7\hat{j} + 3\hat{k}) = 2\mu\hat{i} + (7\mu + 1)\hat{j} + (3\mu + 1)\hat{k}$.To find the point of intersection,equate the components:$1) \lambda - 1 = 2\mu$$2) 2\lambda + 2 = 7\mu + 1 \Rightarrow 2\lambda - 7\mu = -1$$3) \lambda + 1 = 3\mu + 1 \Rightarrow \lambda = 3\mu$Substituting $\lambda = 3\mu$ into the first equation: $3\mu - 1 = 2\mu \Rightarrow \mu = 1$. Then $\lambda = 3(1) = 3$.Substituting $\lambda = 3$ into $L_1$: $\overrightarrow{r} = (3-1)\hat{i} + (2(3)+2)\hat{j} + (3+1)\hat{k} = 2\hat{i} + 8\hat{j} + 4\hat{k}$.The direction vector of $L_3$ is $\overrightarrow{a} + \overrightarrow{b} = (1+2)\hat{i} + (2+7)\hat{j} + (1+3)\hat{k} = 3\hat{i} + 9\hat{j} + 4\hat{k}$.The equation of $L_3$ is $\overrightarrow{r} = (2\hat{i} + 8\hat{j} + 4\hat{k}) + t(3\hat{i} + 9\hat{j} + 4\hat{k})$.For $t = 2$,$\overrightarrow{r} = (2+6)\hat{i} + (8+18)\hat{j} + (4+8)\hat{k} = 8\hat{i} + 26\hat{j} + 12\hat{k}$.Thus,the line $L_3$ passes through the point $(8, 26, 12)$.

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