JEE Mathematics 2026 Question with Solution

Given: $L$ is the line $\frac{x+1}{2} = \frac{y+1}{3} = \frac{z+3}{6}$. The required points on $L$ are at distance $7$ along $L$ from the line $\frac{x+1}{2} = \frac{y+1}{3} = \frac{z-9}{0}$.

Find: $\sum_{(a,b,c)\in S}(a+b+c)$.

Write the parametric form of $L$:

$$\frac{x+1}{2} = \frac{y+1}{3} = \frac{z+3}{6} = t$$

So,

$$x = 2t - 1, \quad y = 3t - 1, \quad z = 6t - 3$$

The second line is interpreted as

$$\frac{x+1}{2} = \frac{y+1}{3}, \quad z = 9$$

Substitute $z = 9$ into the parametric form of $L$:

$$6t - 3 = 9$$ $$t = 2$$

Hence the intersection point is

$$P(3,5,9)$$

Direction ratios of $L$ are $(2,3,6)$, whose magnitude is

$$\sqrt{2^2 + 3^2 + 6^2} = 7$$

Therefore, a unit direction vector of $L$ is

$$\left(\frac{2}{7}, \frac{3}{7}, \frac{6}{7}\right)$$

Points at distance $7$ from $P$ along $L$ are

$$P \pm 7\left(\frac{2}{7}, \frac{3}{7}, \frac{6}{7}\right)$$

So the two points are

$$P_1 = (3,5,9) + (2,3,6) = (5,8,15)$$

and

$$P_2 = (3,5,9) - (2,3,6) = (1,2,3)$$

Now,

$$(a+b+c)_{P_1} = 5 + 8 + 15 = 28$$ $$(a+b+c)_{P_2} = 1 + 2 + 3 = 6$$

Therefore,

$$\sum_{(a,b,c)\in S}(a+b+c) = 28 + 6 = 40$$

So, the correct option is B.

Given: The line $L$ has direction ratios $(2,3,6)$ and the reference point on $L$ comes from its intersection with the line having $z = 9$.

Find: The sum of $a+b+c$ over all required points.

Since on $L$,

$$z = 6t - 3$$

putting $z = 9$ gives

$$t = 2$$

and hence

$$P = (3,5,9)$$

Now the direction vector $(2,3,6)$ already has magnitude

$$\sqrt{4+9+36} = 7$$

So moving a distance $7$ along $L$ means moving exactly by $\pm(2,3,6)$. Thus the points are

$$(3,5,9) \pm (2,3,6)$$

which gives $(5,8,15)$ and $(1,2,3)$. Their coordinate sums are $28$ and $6$ respectively, so the total is

$$28 + 6 = 40$$

Therefore, the correct option is B.