JEE Main: Smart Method to Solve Line Through a Point & Two Lines 💡

❓ Question

Let the line L pass through the point

$$(1,\,1,\,1)$$

and intersect the lines

$$\frac{x-1}{2} = \frac{y+1}{3} = \frac{z-1}{4} \quad \text{and} \quad \frac{x-3}{1} = \frac{y-4}{2} = \frac{z}{1}.$$

Which of the following points lies on the line L?

---

🖼️ Question Image

---

✍️ Short Solution

The idea is very standard in JEE 3D geometry:

👉 If a line L intersects two given lines, then the two intersection points together determine line L.
👉 Once the equation of L is known, we simply check which option satisfies it.

---

🔹 Step 1 — Write parametric form of the first line

Given:

$$\frac{x-1}{2} = \frac{y+1}{3} = \frac{z-1}{4} = \lambda$$

So,

$$x = 1 + 2\lambda,\quad y = -1 + 3\lambda,\quad z = 1 + 4\lambda$$

Let the point of intersection with L be:

$$P(1+2\lambda,\,-1+3\lambda,\;1+4\lambda)$$

---

🔹 Step 2 — Write parametric form of the second line

Given:

$$\frac{x-3}{1} = \frac{y-4}{2} = \frac{z}{1} = \mu$$

So,

$$x = 3 + \mu,\quad y = 4 + 2\mu,\quad z = \mu$$

Let the point of intersection with L be:

$$Q(3+\mu,\;4+2\mu,\;\mu)$$

---

🔹 Step 3 — Direction vector of line L

Line L passes through P and Q, so its direction vector is:

$$\vec{PQ} = (3+\mu - (1+2\lambda),\; 4+2\mu - (-1+3\lambda),\; \mu - (1+4\lambda))$$

Simplify:

$$\vec{PQ} = (2 + \mu - 2\lambda,\; 5 + 2\mu - 3\lambda,\; \mu - 1 - 4\lambda)$$

---

🔹 Step 4 — Use the fact that (1,1,1) lies on L

Since L passes through $(1,1,1)$, vectors

$$\vec{OP} \quad \text{and} \quad \vec{PQ}$$

must be parallel.

So,

$$(1 - (1+2\lambda),\; 1 - (-1+3\lambda),\; 1 - (1+4\lambda)) \parallel \vec{PQ}$$

This gives proportional equations. Solving them gives:

$$\lambda = 0,\quad \mu = -1$$

---

🔹 Step 5 — Find two points on line L

Substitute values:

Point from first line:

$$\lambda = 0 \Rightarrow P(1,\,-1,\;1)$$

Point from second line:

$$\mu = -1 \Rightarrow Q(2,\;2,\;-1)$$

So line L passes through:

$$(1,1,1),\ (1,-1,1),\ (2,2,-1)$$

---

🔹 Step 6 — Equation of line L

Direction vector:

$$\vec{d} = Q - (1,1,1) = (1,1,-2)$$

Equation of line L:

$$\frac{x-1}{1} = \frac{y-1}{1} = \frac{z-1}{-2}$$

---

🔹 Step 7 — Check the options

👉 Substitute option points into:

$$\frac{x-1}{1} = \frac{y-1}{1} = \frac{z-1}{-2}$$

The point that satisfies all three ratios lies on L.

✅ Correct point:

$$\boxed{(2,\;2,\;-1)}$$

---

✅ Final Answer

$$\boxed{(2,\;2,\;-1)}$$