JEE Main 2024 — 27 January, Shift 2 — Question 29

Step 1: Find the intersection point $P$ of the lines $L_1$ and $L_2$.** The first line $L_1$ is given by:

$$\frac{x-2}{2} = \frac{y}{-2} = \frac{z-7}{16} = \lambda$$

Any point on $L_1$ can be represented as $P_1(\lambda) = (2+2\lambda, -2\lambda, 7+16\lambda)$.

The second line $L_2$ is given by:

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

Any point on $L_2$ can be represented as $P_2(\mu) = (-3+4\mu, -2+3\mu, -2+\mu)$.

For the lines to intersect at point $P$, the coordinates must be equal:

1. $2+2\lambda = -3+4\mu \quad \Rightarrow \quad 2\lambda - 4\mu = -5 \quad \text{(Equation 1)}$
2. $-2\lambda = -2+3\mu \quad \Rightarrow \quad -2\lambda - 3\mu = -2 \quad \text{(Equation 2)}$
3. $7+16\lambda = -2+\mu \quad \Rightarrow \quad 16\lambda - \mu = -9 \quad \text{(Equation 3)}$

Add Equation 1 and Equation 2: $(2\lambda - 4\mu) + (-2\lambda - 3\mu) = -5 + (-2)$ $-7\mu = -7$

$$\mu = 1$$

Substitute $\mu = 1$ into Equation 2: $-2\lambda - 3(1) = -2$ $-2\lambda - 3 = -2$ $-2\lambda = 1$

$$\lambda = -\frac{1}{2}$$

Now, verify these values in Equation 3: $16\left(-\frac{1}{2}\right) - (1) = -8 - 1 = -9$. The values of $\lambda = -\frac{1}{2}$ and $\mu = 1$ satisfy all three equations, confirming the intersection.

Substitute $\lambda = -\frac{1}{2}$ into the parametric form of $P_1(\lambda)$ to find point $P$: $P = \left(2+2\left(-\frac{1}{2}\right), -2\left(-\frac{1}{2}\right), 7+16\left(-\frac{1}{2}\right)\right)$ $P = (2-1, 1, 7-8)$

$$P = (1, 1, -1)$$

Step 2: Find the distance $l$ of point $P$ from the line $L_3$.** The third line $L_3$ is given by:

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

A point on $L_3$ is $A = (-1, 1, 1)$. The direction vector of $L_3$ is $\vec{b} = (2, 3, 1)$.

The vector from point $A$ on the line $L_3$ to point $P$ is $\vec{AP}$:

$$\vec{AP} = P - A = (1 - (-1), 1 - 1, -1 - 1) = (2, 0, -2)$$

The distance $l$ from point $P$ to line $L_3$ is given by the formula:

$$l = \frac{|\vec{AP} \times \vec{b}|}{|\vec{b}|}$$

First, calculate the cross product $\vec{AP} \times \vec{b}$:

$$\vec{AP} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 0 & -2 \\ 2 & 3 & 1 \end{vmatrix}$$ $$= \hat{i}((0)(1) - (-2)(3)) - \hat{j}((2)(1) - (-2)(2)) + \hat{k}((2)(3) - (0)(2))$$ $$= \hat{i}(0 + 6) - \hat{j}(2 + 4) + \hat{k}(6 - 0)$$ $$= 6\hat{i} - 6\hat{j} + 6\hat{k}$$

Next, calculate the magnitude of the cross product:

$$|\vec{AP} \times \vec{b}| = \sqrt{6^2 + (-6)^2 + 6^2} = \sqrt{36 + 36 + 36} = \sqrt{3 \times 36} = 6\sqrt{3}$$

Next, calculate the magnitude of the direction vector $\vec{b}$:

$$|\vec{b}| = \sqrt{2^2 + 3^2 + 1^2} = \sqrt{4 + 9 + 1} = \sqrt{14}$$

Now, calculate the distance $l$:

$$l = \frac{6\sqrt{3}}{\sqrt{14}}$$

Step 3: Calculate $14l^2$.** First, calculate $l^2$:

$$l^2 = \left(\frac{6\sqrt{3}}{\sqrt{14}}\right)^2 = \frac{(6\sqrt{3})^2}{(\sqrt{14})^2} = \frac{36 \times 3}{14} = \frac{108}{14} = \frac{54}{7}$$

Finally, calculate $14l^2$:

$$14l^2 = 14 \times \frac{54}{7} = 2 \times 54 = 108$$

The final answer is $\boxed{108}$.