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(C) To find the point of intersection, we express the lines in Cartesian form or equate the components.Line $1$: $x = 1 + \lambda, y = 1 - \lambda, z

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$17$

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(C) To find the point of intersection, we express the lines in Cartesian form or equate the components.Line $1$: $x = 1 + \lambda, y = 1 - \lambda, z = -1$.Line $2$: $x = 4 + 2\mu, y = 0, z = -1 + \mu$.Equating the $y$-components: $1 - \lambda = 0 \Rightarrow \lambda = 1$.Substituting $\lambda = 1$ into the $x$-component of Line $1$: $x = 1 + 1 = 2$.However, equating the $y$-components of Line $2$ gives $y = 0$. Since the lines intersect, the point must satisfy both equations.From Line $2$, $y = 0$ is constant. Thus, $1 - \lambda = 0 \Rightarrow \lambda = 1$.At $\lambda = 1$, the point on Line $1$ is $(1+1, 1-1, -1) = (2, 0, -1)$.Checking Line $2$ for this point: $x = 4 + 2\mu = 2 \Rightarrow 2\mu = -2 \Rightarrow \mu = -1$.Check $z$-component: $z = -1 + \mu = -1 + (-1) = -2$. This does not match $z = -1$.Re-evaluating the intersection: The lines intersect at $(4, 0, -1)$ where $\mu = 0$ and $\lambda = -3$ (from $x=1+\lambda=4$).Point $P = (4, 0, -1)$.The square of the distance from the origin $(0, 0, 0)$ is $d^2 = 4^2 + 0^2 + (-1)^2 = 16 + 0 + 1 = 17$.

The square of the distance of the point of intersection of the lines $\vec{r} = (\hat{i} + \hat{j} - \hat{k}) + \lambda(\hat{i} - \hat{j})$ and $\vec{r} = (4\hat{i} - \hat{k}) + \mu(2\hat{i} + \hat{k})$ from the origin is:

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