The distance of the point $(1,-2,3)$ from the plane $x-y+z=5$ measured parallel to the line $\frac{\mathrm{x}}{2}=\frac{\mathrm{y}}{3}=\frac{\mathrm{z}}{-6}$ is:

JEE Main · 2020 · 04 Sep 2020 · Shift 2 · Three Dimensional Geometry

The distance of the point

(1, - 2, 3)

from the plane

x - y + z = 5

measured parallel to the line

\frac{x}{2} = \frac{y}{3} = \frac{z}{- 6}

is:

(A) $\frac{1}{7}$
(B)7
(C) $\frac{7}{5}$
(D)1✓ Correct

Answer: (D)

Solution

A line through

P (1, - 2, 3)

parallel to the given line is

\frac{x - 1}{2} = \frac{y + 2}{3} = \frac{z - 3}{- 6} = λ

, so a general point on it is

(2 λ + 1, 3 λ - 2, - 6 λ + 3)

. Requiring this to lie on the plane

x - y + z = 5

gives

(2 λ + 1) - (3 λ - 2) + (- 6 λ + 3) = 5

, so

- 7 λ + 6 = 5

and

λ = - \frac{1}{7}

. The distance is then

∣ λ ∣ 2^{2} + 3^{2} + 6^{2} = \frac{1}{7} \cdot 7 = 1

JEE Main · 2020 · 04 Sep 2020 · Shift 2 · Single correct option

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