Find Distance Squared of Intersection Point of Lines from Origin

To find the square of the distance of the point of intersection of the given lines from the origin, we first express both line equations in parametric form.

The equation of the first line $L_1$ is: $\vec{r} = (1 + a\lambda)\hat{i} + (1 - \lambda)\hat{j} - \hat{k}$

So, any point on line $L_1$ has coordinates: $x_1 = 1 + a\lambda, \quad y_1 = 1 - \lambda, \quad z_1 = -1$

The equation of the second line $L_2$ is: $\vec{r} = (4 + 2\mu)\hat{i} + 0\hat{j} + (-1 + a\mu)\hat{k}$

So, any point on line $L_2$ has coordinates: $x_2 = 4 + 2\mu, \quad y_2 = 0, \quad z_2 = -1 + a\mu$

Since the two lines intersect, their corresponding coordinates at the point of intersection must be equal:

1. Equating $y$-coordinates: $1 - \lambda = 0 \implies \lambda = 1$

2. Equating $z$-coordinates: $-1 = -1 + a\mu \implies a\mu = 0$ Since it is given that $a \neq 0$, we must have: $\mu = 0$

3. Equating $x$-coordinates: $1 + a\lambda = 4 + 2\mu$

Substitute $\lambda = 1$ and $\mu = 0$ into the $x$-coordinate equation: $1 + a(1) = 4 + 2(0)$ $1 + a = 4 \implies a = 3$

Now, substitute $\lambda = 1$ and $a = 3$ back into the coordinates of $L_1$ (or $\mu = 0$ into $L_2$) to find the point of intersection $P$: $P = (1 + 3(1), \, 1 - 1, \, -1) = (4, 0, -1)$

The square of the distance of $P(4, 0, -1)$ from the origin $O(0, 0, 0)$ is: $OP^2 = 4^2 + 0^2 + (-1)^2 = 16 + 0 + 1 = 17$

Therefore, the correct option is C.