JEE Mathematics 2026 Question with Solution

Given: $P(\alpha,\beta,\gamma)$ lies on $\frac{x-1}{2} = \frac{y+1}{-3} = z$ and is at distance $4\sqrt{14}$ from $(1,-1,0)$, nearer to the origin.

Find: The shortest distance between the lines $\frac{x-\alpha}{1} = \frac{y-\beta}{2} = \frac{z-\gamma}{3}$ and $\frac{x+5}{2} = \frac{y-10}{1} = \frac{z-3}{1}$.

From

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

we get

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

The distance from $(1,-1,0)$ to $P$ is

$$\sqrt{(2t)^2 + (-3t)^2 + t^2} = 4\sqrt{14}$$

So,

$$\sqrt{14t^2} = 4\sqrt{14}$$

which gives

$$|t| = 4$$

Since the point is nearer to the origin, take $t = -4$. Therefore,

$$\alpha = -7, \qquad \beta = 11, \qquad \gamma = -4$$

Now the first line passes through $P_1(-7,11,-4)$ with direction vector

$$\vec{d}_1 = \langle 1,2,3 \rangle$$

The second line passes through $P_2(-5,10,3)$ with direction vector

$$\vec{d}_2 = \langle 2,1,1 \rangle$$

For two skew lines, shortest distance is

$$\frac{|(\vec{P_2P_1}) \cdot (\vec{d}_1 \times \vec{d}_2)|}{|\vec{d}_1 \times \vec{d}_2|}$$

Here,

$$\vec{P_2P_1} = \langle -2,1,-7 \rangle$$

Also,

$$\vec{d}_1 \times \vec{d}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 2 & 1 & 1 \end{vmatrix} = \langle -1,5,-3 \rangle$$

Hence,

$$|\vec{d}_1 \times \vec{d}_2| = \sqrt{(-1)^2 + 5^2 + (-3)^2} = \sqrt{35}$$

and

$$|(\vec{P_2P_1}) \cdot (\vec{d}_1 \times \vec{d}_2)| = |(-2)(-1) + (1)(5) + (-7)(-3)| = 28$$

Therefore,

$$\text{Shortest distance} = \frac{28}{\sqrt{35}} = 4\sqrt{\frac{7}{5}}$$

So, the correct option is D.

Given: $P$ is on the line $\frac{x-1}{2} = \frac{y+1}{-3} = z$ and its distance from $(1,-1,0)$ is $4\sqrt{14}$.

Find: The shortest distance between the two given lines.

Write the first given line in parametric form:

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

So the point $P$ is

$$P = (1+2t, -1-3t, t)$$

Distance from $(1,-1,0)$ is

$$\sqrt{(1+2t-1)^2 + (-1-3t+1)^2 + (t-0)^2} = 4\sqrt{14}$$

That is,

$$\sqrt{(2t)^2 + (-3t)^2 + t^2} = 4\sqrt{14}$$ $$\sqrt{14t^2} = 4\sqrt{14}$$ $$|t| = 4$$

The two possible points are obtained from $t = 4$ and $t = -4$. Since the required point is nearer to the origin, choose $t = -4$. Thus,

$$P = (-7,11,-4)$$

Now the first required line is

$$\frac{x+7}{1} = \frac{y-11}{2} = \frac{z+4}{3}$$

so a point on it is $P_1(-7,11,-4)$ and direction vector is

$$\vec{d}_1 = \langle 1,2,3 \rangle$$

For the second line,

$$\frac{x+5}{2} = \frac{y-10}{1} = \frac{z-3}{1}$$

so a point on it is $P_2(-5,10,3)$ and direction vector is

$$\vec{d}_2 = \langle 2,1,1 \rangle$$

Take the joining vector:

$$\vec{P_2P_1} = P_1 - P_2 = \langle -2,1,-7 \rangle$$

Now compute the cross product:

$$\vec{d}_1 \times \vec{d}_2 = \langle -1,5,-3 \rangle$$

Its magnitude is

$$\sqrt{1+25+9} = \sqrt{35}$$

The scalar triple product magnitude is

$$|\vec{P_2P_1} \cdot (\vec{d}_1 \times \vec{d}_2)| = |(-2)(-1) + (1)(5) + (-7)(-3)| = 28$$

Hence the shortest distance is

$$\frac{28}{\sqrt{35}} = 4\sqrt{\frac{7}{5}}$$

Therefore, the shortest distance is $4\sqrt{\frac{7}{5}}$ and the correct option is D.