If the distance of the point \(\mathrm{P}(43, \alpha, \beta), \beta<0\), from the line \(\overrightarrow{\mathrm{r}}=4 \hat{\mathrm{i}}-\hat{\mathrm{k}}+\mu(2 \hat{\mathrm{i}}+3 \hat{\mathrm{k}}), \mu \in \mathbf{R}\) along a line with direction ratios \(3,-1,0\) is \(13 \sqrt{10}\), then \(\alpha^{2}+\beta^{2}\) is equal to ______。

JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Three Dimensional Geometry

If the distance of the point

P (43, α, β), β < 0

, from the line

r = 4 \hat{i} - \hat{k} + μ (2 \hat{i} + 3 \hat{k}), μ \in R

along a line with direction ratios

3, - 1, 0

1310

, then

α^{2} + β^{2}

is equal to ______。

Answer: 170

The line through

P (43, α, β)

with direction

(3, - 1, 0)

(43 + 3 λ, α - λ, β)

, and it must meet

r = (4, 0, - 1) + μ (2, 0, 3)

. Every point of the given line has

y

-coordinate

0

, so

α - λ = 0

, i.e.

λ = α

. The travelled distance is

∣ λ ∣ 3^{2} + (- 1)^{2} + 0^{2} = ∣ λ ∣ 10 = 1310

, so

∣ λ ∣ = 13

. Matching the

x

- and

z

-coordinates,

43 + 3 λ = 4 + 2 μ

gives

μ = \frac{39 + 3 λ}{2}

, and

β = - 1 + 3 μ = \frac{115 + 9 λ}{2}

. Taking

λ = 13

gives

β = 116 > 0

, which is rejected; taking

λ = - 13

gives

β = - 1 < 0

, as required, with

α = - 13

. Hence

α^{2} + β^{2} = 169 + 1 = 170

JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Integer answer