JEE Mathematics 2025 Question with Solution

Given:

• $L_1: \frac{x-3}{3}=\frac{y-\alpha}{-1}=\frac{z-3}{1}$
• $L_2: \frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-\beta}{4}$
• Shortest distance between the lines is $3\sqrt{30}$

Find: The positive value of $5\alpha+\beta$.

For skew lines,

$$d=\frac{\left|(\mathbf{r}_2-\mathbf{r}_1)\cdot(\mathbf{a}_1\times\mathbf{a}_2)\right|}{\left|\mathbf{a}_1\times\mathbf{a}_2\right|}$$

Here,

$$\mathbf{r}_1=(3,\alpha,3),\quad \mathbf{a}_1=\langle 3,-1,1\rangle$$ $$\mathbf{r}_2=(-3,-7,\beta),\quad \mathbf{a}_2=\langle -3,2,4\rangle$$

Now,

$$\mathbf{a}_1\times\mathbf{a}_2=\begin{vmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\\ 3&-1&1\\ -3&2&4 \end{vmatrix}=\langle -6,-15,3\rangle$$

Therefore,

$$\left|\mathbf{a}_1\times\mathbf{a}_2\right|=\sqrt{36+225+9}=3\sqrt{30}$$

Also,

$$\mathbf{r}_2-\mathbf{r}_1=\langle -6,-7-\alpha,\beta-3\rangle$$

So the scalar triple product is

$$(\mathbf{r}_2-\mathbf{r}_1)\cdot(\mathbf{a}_1\times\mathbf{a}_2)=(-6)(-6)+(-7-\alpha)(-15)+(\beta-3)(3)$$ $$=36+105+15\alpha+3\beta-9=132+15\alpha+3\beta$$

Using the given distance,

$$\frac{\left|132+15\alpha+3\beta\right|}{3\sqrt{30}}=3\sqrt{30}$$

Hence,

$$\left|132+15\alpha+3\beta\right|=(3\sqrt{30})^2=270$$ $$15\alpha+3\beta=\pm 138$$ $$5\alpha+\beta=\pm 46$$

The positive value is $46$. Therefore, the correct option is B.

Given: The two lines are in symmetric form.

Find: The positive value of $5\alpha+\beta$.

Take point $A(3,\alpha,3)$ on the first line and point $B(-3,-7,\beta)$ on the second line. Their direction vectors are

$$\vec{p}=3\hat{i}-\hat{j}+\hat{k},\qquad \vec{q}=-3\hat{i}+2\hat{j}+4\hat{k}$$

Now,

$$\vec{p}\times\vec{q}=6\hat{i}+15\hat{j}-9\hat{k}$$

whose magnitude is

$$\sqrt{6^2+15^2+(-9)^2}=\sqrt{36+225+81}=\sqrt{342}=3\sqrt{38}$$

the solution shows this intermediate cross-product, but the consistent computation from the same working gives magnitude $3\sqrt{30}$ when using the equivalent vector $\langle -6,-15,3\rangle$ from the determinant expansion in the detailed method. Using the consistent determinant-based result,

$$\left|\vec{p}\times\vec{q}\right|=3\sqrt{30}$$

Also,

$$\vec{BA}=6\hat{i}+(\alpha+7)\hat{j}+(3-\beta)\hat{k}$$

Then,

$$\frac{|\vec{BA}\cdot(\vec{p}\times\vec{q})|}{|\vec{p}\times\vec{q}|}=3\sqrt{30}$$

which leads to

$$15\alpha+3\beta=138$$

Therefore,

$$5\alpha+\beta=46$$

Thus the correct option is B. Note that one approach in the source has an inconsistent cross-product entry, but both the final detailed computation and the listed correct option support $46$.