Let PQR be a triangle such that $\overrightarrow{\mathrm{PQ}}=-2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{PR}}=a \hat{\mathrm{i}}+b \hat{\mathrm{j}}-4 \hat{\mathrm{k}}, a, b \in \mathbb{Z}$. Let $S$ be the point on QR, which is equidistant from the lines PQ and PR. If $|\overrightarrow{\mathrm{PR}}|=9$ and $\overrightarrow{\mathrm{PS}}=\hat{\mathrm{i}}-7 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}$, then the value of $3 a-4 b$ is

JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Vector Algebra

Let PQR be a triangle such that

PQ = - 2 \hat{i} - \hat{j} + 2 \hat{k}

and

PR = a \hat{i} + b \hat{j} - 4 \hat{k}, a, b \in Z

. Let

S

be the point on QR, which is equidistant from the lines PQ and PR. If

∣ PR ∣ = 9

and

PS = \hat{i} - 7 \hat{j} + 2 \hat{k}

, then the value of

3 a - 4 b

Answer: 37

Since

∣ P R ∣ = 9

a^{2} + b^{2} + 16 = 81

, giving

a^{2} + b^{2} = 65

. The point

S

QR

equidistant from the lines

P Q

and

P R

makes

P S

the internal bisector of

∠ QP R

, so

P S

is parallel to the sum of the unit vectors along

P Q

and

P R

. With

∣ P Q ∣ = 3

and

∣ P R ∣ = 9

, this direction is proportional to

9 (\frac{( - 2 , - 1 , 2 )}{3} + \frac{( a , b , - 4 )}{9}) = (a - 6, b - 3, 2)

. Matching with

P S = (1, - 7, 2)

, the equal third components fix the factor as

1

, so

a - 6 = 1

and

b - 3 = - 7

, giving

a = 7

and

b = - 4

, which satisfy

a^{2} + b^{2} = 65

. Hence

3 a - 4 b = 21 + 16 = 37

JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Integer answer