JEE Main · 2026 · 02 Apr 2026 · Shift 1 · Three Dimensional Geometry

If the point of intersection of the lines

\frac{x + 1}{3} = \frac{y + a}{5} = \frac{z + b + 1}{7}

and

\frac{x - 2}{1} = \frac{y - b}{4} = \frac{z - 2 a}{7}

lies on

x y

-plane, then the value of

a + b

is :

(A)2
(B)5
(C)7✓ Correct
(D)9

Answer: (C)

Solution

Points on the two lines are

(3 t - 1, 5 t - a, 7 t - b - 1)

and

(s + 2, 4 s + b, 7 s + 2 a)

. Lying on the

x y

-plane forces

z = 0

on each, so

t = \frac{b + 1}{7}

and

s = - \frac{2 a}{7}

. Equating

x

-coordinates,

3 t - 1 = s + 2 \Rightarrow s = 3 t - 3

, which gives

- \frac{2 a}{7} = 3 \cdot \frac{b + 1}{7} - 3

, i.e.

2 a + 3 b = 18

. Equating

y

-coordinates,

5 t - a = 4 s + b

with

s = 3 t - 3

and

7 t = b + 1

gives

a = 11 - 2 b

. Solving

2 a + 3 b = 18

together with

a = 11 - 2 b

yields

b = 4

and

a = 3

, so

a + b = 7

JEE Main · 2026 · 02 Apr 2026 · Shift 1 · Single correct option

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