Question

Let a line \(L_1\) pass through the origin and be perpendicular to the lines \(L_2: \vec{r} = (3 + t)\hat{i} + (2t - 1)\hat{j} + (2t + 4)\hat{k}\) and \(L_3: \vec{r} = (3 + 2s)\hat{i} + (3 + 2s)\hat{j} + (2 + s)\hat{k}\). If \((a, b, c)\), \(a \in \mathbb{Z}\), is the point on \(L_3\) at a distance of \(\sqrt{17}\) from the point of intersection of \(L_1\) and \(L_2\), then \((a + b + c)^2\) is equal to ________.

Answer 1

Correct Answer: 4

Step 1: Understanding the Concept:
We first find the direction ratios of \(L_1\) by taking the cross product of the directions of \(L_2\) and \(L_3\). Then, find the intersection point of \(L_1\) and \(L_2\), and finally use the distance formula to find the point on \(L_3\).

Step 2: Key Formula or Approach:
1. Direction of \(L_1 = \vec{d_2} \times \vec{d_3}\).
2. \(\vec{d_2} = (1,2,2)\), \(\vec{d_3} = (2,2,1)\).

Step 3: Detailed Explanation:
1. Direction of \(L_1\):
\[ \vec{d_1} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & 2 \\ 2 & 2 & 1 \end{vmatrix} \] \[ = \mathbf{i}(2 - 4) - \mathbf{j}(1 - 4) + \mathbf{k}(2 - 4) = (-2, 3, -2) \] 2. Intersection of \(L_1\) and \(L_2\):
Line \(L_1\): \[ \vec{r} = k(-2, 3, -2) \] Equating with \(L_2\): \[ 3 + t = -2k,\quad 2t - 1 = 3k,\quad 2t + 4 = -2k \] From: \[ 3 + t = 2t + 4 \Rightarrow t = -1 \] Point of intersection: \[ P = (3 - 1,\; 2(-1) - 1,\; 2(-1) + 4) = (2, -3, 2) \] 3. Point on \(L_3\):
General point: \[ Q = (3 + 2s,\; 3 + 2s,\; 2 + s) \] Using distance condition \(PQ^2 = 17\): \[ (3 + 2s - 2)^2 + (3 + 2s + 3)^2 + (2 + s - 2)^2 = 17 \] \[ (2s + 1)^2 + (2s + 6)^2 + s^2 = 17 \] \[ 4s^2 + 4s + 1 + 4s^2 + 24s + 36 + s^2 = 17 \] \[ 9s^2 + 28s + 20 = 0 \] \[ (s + 2)(9s + 10) = 0 \] \[ s = -2 \quad \text{or} \quad s = -\frac{10}{9} \] Since integer value is required: \[ s = -2 \] \[ Q = (3 - 4,\; 3 - 4,\; 2 - 2) = (-1, -1, 0) \] \[ a = -1,\quad b = -1,\quad c = 0 \] 
Step 4: Final Answer:
\[ (a + b + c)^2 = (-1 - 1 + 0)^2 = (-2)^2 = 4 \]