Two friends are planning a road trip. One friend stays in City A represented by the position vector…

Let $\vec{a} = -2\hat{i} + 3\hat{j} + 5\hat{k}$, $\vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}$, $\vec{c} = 7\hat{i} - \hat{k}$. The vector equation of AB is given by $\vec{r} = \vec{a} + \lambda(\vec{b} - \vec{a})$. $\vec{r} = (-2\hat{i} + 3\hat{j} + 5\hat{k}) + \lambda[(\hat{i} + 2\hat{j} + 3\hat{k}) - (-2\hat{i} + 3\hat{j} + 5\hat{k})]$ $\vec{r} = (-2\hat{i} + 3\hat{j} + 5\hat{k}) + \lambda(3\hat{i} - \hat{j} - 2\hat{k})$

The three cities will lie on the same straight path if they are collinear. So if C lies on AB, the three points are collinear. When $7\hat{i} - \hat{k} = (-2\hat{i} + 3\hat{j} + 5\hat{k}) + \lambda(3\hat{i} - \hat{j} - 2\hat{k})$ we get, $7 = -2 + 3\lambda$, $0 = 3 - \lambda$, $-1 = 5 - 2\lambda$. The value of $\lambda = 3$ satisfies all three equations. So, C lies on AB. Hence, we can conclude that they three cities lie on the same straight path.

Let P be the destination point. Co-ordinates of P is $(3r - 2, -r + 3, -2r + 5)$. $CP = \sqrt{126} = \sqrt{(3r-2-7)^2 + (3-r)^2 + (5-2r+1)^2}$ $r(r - 6) = 0$, $r = 0$ or $6$ If $r = 0$, P is $(-2, 3, 5)$ which is the City A. So, for $r = 6$, P is $(16, -3, -7)$ Position vector of P is $16\hat{i} - 3\hat{j} - 7\hat{k}$.