JEE Mathematics 2025 Question with Solution

Given: A line passes through $(-1, 2, 1)$ and is parallel to $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z}{4}$. It intersects the line $\frac{x + 2}{3} = \frac{y - 3}{2} = \frac{z - 4}{1}$ at $P$.

Find: The distance between $P$ and $Q(4, -5, 1)$.

The direction ratios of the first given line are $\langle 2, 3, 4 \rangle$. Therefore, the required line through $(-1, 2, 1)$ has parametric equations

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

The second line can be written as

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

so its parametric equations are

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

At the point of intersection, coordinates from both lines are equal. So,

$$-1 + 2t = 3\lambda - 2$$ $$2 + 3t = 2\lambda + 3$$ $$1 + 4t = \lambda + 4$$

From the first equation,

$$2t + 1 = 3\lambda$$

so

$$\lambda = \frac{2t + 1}{3}$$

From the second equation,

$$3t - 1 = 2\lambda$$

so

$$\lambda = \frac{3t - 1}{2}$$

Equating these,

$$\frac{2t + 1}{3} = \frac{3t - 1}{2}$$ $$4t + 2 = 9t - 3$$ $$5t = 5$$ $$t = 1$$

Substituting $t = 1$,

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

Now substitute $t = 1$ into the first line:

$$x = -1 + 2(1) = 1, \quad y = 2 + 3(1) = 5, \quad z = 1 + 4(1) = 5$$

Hence,

$$P = (1, 5, 5)$$

Now use the distance formula between $P(1, 5, 5)$ and $Q(4, -5, 1)$:

$$d = \sqrt{(4 - 1)^2 + (-5 - 5)^2 + (1 - 5)^2}$$ $$d = \sqrt{3^2 + (-10)^2 + (-4)^2}$$ $$d = \sqrt{9 + 100 + 16}$$ $$d = \sqrt{125} = 5\sqrt{5}$$

Therefore, the distance of $P$ from $Q$ is $5\sqrt{5}$. The correct option is D.

Given: The first line passes through $(-1,2,1)$ and is parallel to direction ratios $\langle 2,3,4 \rangle$. The second line is $\frac{x+2}{3}=\frac{y-3}{2}=\frac{z-4}{1}$.

Find: The distance from the intersection point $P$ to $Q(4,-5,1)$.

Write the first line as

$$\frac{x-(-1)}{2}=\frac{y-2}{3}=\frac{z-1}{4}=t$$

Hence,

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

Write the second line as

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

Hence,

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

For intersection,

$$-1+2t=-2+3s$$ $$2+3t=3+2s$$ $$1+4t=4+s$$

These become

$$2t-3s=-1$$ $$3t-2s=1$$ $$4t-s=3$$

From

$$4t-s=3$$

we get

$$s=4t-3$$

Substitute into

$$2t-3s=-1$$ $$2t-3(4t-3)=-1$$ $$2t-12t+9=-1$$ $$-10t=-10$$ $$t=1$$

Then,

$$s=4(1)-3=1$$

Now,

$$P=( -1+2,\; 2+3,\; 1+4 )=(1,5,5)$$

Distance from $P(1,5,5)$ to $Q(4,-5,1)$ is

$$\sqrt{(4-1)^2+(-5-5)^2+(1-5)^2}$$ $$=\sqrt{9+100+16}$$ $$=\sqrt{125}=5\sqrt{5}$$

Therefore, the required distance is $5\sqrt{5}$ and the correct option is D.