JEE Mathematics 2026 Question with Solution

Given: Line $L_1$ is parallel to $(-3, 2, 4)$ and passes through $(2, 6, 7)$. Line $L_2$ is parallel to $(2, 1, 3)$ and passes through $(4, 3, 5)$. Line $L_3$ is parallel to $(-3, 5, 16)$ and intersects $L_1$ and $L_2$ at points $C$ and $D$.

Find: $|\vec{CD}|^2$.

Take a general point $C$ on $L_1$ and a general point $D$ on $L_2$.

$$\vec{c} = (2 - 3\lambda, 6 + 2\lambda, 7 + 4\lambda)$$ $$\vec{d} = (4 + 2\mu, 3 + \mu, 5 + 3\mu)$$

Then

$$\vec{CD} = \vec{d} - \vec{c} = (2 + 2\mu + 3\lambda, -3 + \mu - 2\lambda, -2 + 3\mu - 4\lambda)$$

Since $\vec{CD}$ is parallel to $(-3, 5, 16)$,

$$\frac{2 + 2\mu + 3\lambda}{-3} = \frac{-3 + \mu - 2\lambda}{5} = \frac{-2 + 3\mu - 4\lambda}{16} = k$$

From the first two ratios,

$$10 + 10\mu + 15\lambda = 9 - 3\mu + 6\lambda$$ $$13\mu + 9\lambda = -1$$

From the last two ratios,

$$-48 + 16\mu - 32\lambda = -10 + 15\mu - 20\lambda$$ $$\mu - 12\lambda = 38$$

Solving these equations gives

$$\lambda = -3, \qquad \mu = 2$$

Now,

$$k = \frac{2 + 2(2) + 3(-3)}{-3} = \frac{-3}{-3} = 1$$

Hence,

$$\vec{CD} = 1 \cdot (-3, 5, 16) = (-3, 5, 16)$$

Therefore,

$$|\vec{CD}|^2 = (-3)^2 + 5^2 + 16^2 = 9 + 25 + 256 = 290$$

So, the correct option is A.

Given: The intersection points $C$ and $D$ lie on $L_1$ and $L_2$ respectively, and the line joining them is parallel to $(-3, 5, 16)$.

Find: The value of $|\vec{CD}|^2$.

Write the parametric forms of the two given lines:

$$L_1: (x, y, z) = (2, 6, 7) + \lambda(-3, 2, 4)$$ $$L_2: (x, y, z) = (4, 3, 5) + \mu(2, 1, 3)$$

So,

$$C = (2 - 3\lambda, 6 + 2\lambda, 7 + 4\lambda)$$ $$D = (4 + 2\mu, 3 + \mu, 5 + 3\mu)$$

Now form the vector from $C$ to $D$:

$$\vec{CD} = (4 + 2\mu - (2 - 3\lambda), 3 + \mu - (6 + 2\lambda), 5 + 3\mu - (7 + 4\lambda))$$ $$\vec{CD} = (2 + 2\mu + 3\lambda, -3 + \mu - 2\lambda, -2 + 3\mu - 4\lambda)$$

Because $L_3$ is parallel to $(-3, 5, 16)$, the vector $\vec{CD}$ must be proportional to this direction vector. Thus,

$$(2 + 2\mu + 3\lambda, -3 + \mu - 2\lambda, -2 + 3\mu - 4\lambda) = k(-3, 5, 16)$$

Equating component ratios gives two independent linear equations:

$$13\mu + 9\lambda = -1$$ $$\mu - 12\lambda = 38$$

On solving,

$$\lambda = -3, \qquad \mu = 2$$

Then the proportionality constant is

$$k = 1$$

Hence,

$$\vec{CD} = (-3, 5, 16)$$

Now compute the squared magnitude:

$$|\vec{CD}|^2 = (-3)^2 + 5^2 + 16^2$$ $$= 9 + 25 + 256 = 290$$

Therefore, $|\vec{CD}|^2 = 290$ and the correct option is A.