The square of the distance of the point of intersection of the...

Equating the position vectors of the two lines for intersection:

$(1 + a\lambda)\hat{i} + (1 - \lambda)\hat{j} - \hat{k} = (4 + 2\mu)\hat{i} + 0\hat{j} + (-1 + a\mu)\hat{k}$

Comparing the coefficients of $\hat{i}, \hat{j}, \hat{k}$:

$1 - \lambda = 0 \Rightarrow \lambda = 1$

$-1 = -1 + a\mu \Rightarrow a\mu = 0$

Since $a \neq 0$, we get $\mu = 0$.

Substituting $\lambda = 1$ and $\mu = 0$ in the $\hat{i}$ component equation:

$1 + a(1) = 4 + 2(0) \Rightarrow a = 3$

The position vector of the point of intersection is obtained by substituting $\mu = 0$ in the second line:

$\vec{r} = 4\hat{i} - \hat{k}$

The coordinates of the point of intersection are $(4, 0, -1)$.

The square of the distance from the origin is $4^2 + 0^2 + (-1)^2 = 16 + 1 = 17$.

Answer: $17$