Let the equation of the plane P containing the line x + 10 = (8 - y)/(2) = z be ax + by + 3z = 2 ( a + b ) and the distance of the plane P from the point ( 1, 27, 7 ) be c . Then a ^ 2 + b ^ 2 + c ^ 2 is equal to

JEE Main 202329 JanuaryMorning Shift

Exam: JEE Main
Year: 2023
Date: 29 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Three Dimensional Geometry

JEE Main 2023 · 29 January · Morning Shift · Mathematics · Three Dimensional Geometry

Let the equation of the plane P containing the line x + 10 = (8 -...

Let the equation of the plane

P

containing the line

x + 10 = \dfrac{8 - y}{2} = z

ax + by + 3z = 2 ( a + b )

and the distance of the plane

P

from the point

( 1, 27, 7 )

c

. Then

{a}^{2} + {b}^{2} + {c}^{2}

is equal to

Correct answer

Step-by-step solution

Given,

The line $\dfrac{x + 10}{1} = \dfrac{y - 8}{- 2} = \dfrac{z}{1}$ which passes through the point $( - 10, 8, 0 )$

Also given the equation of plane $ax + by + 3z = 2 ( a + b ) and the point$ ( - 10, 8, 0 ) $will lie on it as plane contains the line,So,$ a\left(- 10\right) + b \times 8 + 3 \times 0 = 2 ( a + b ) $$ ⇒ b = 2a . . . . . . \left(1\right) $Now product of direction ratio of line and direction ratio of plane will be zero as they both are perpendicular,So,$ a \times 1 + b \times \left(- 2\right) + 3 \times 1 = 0 $$ ⇒ a - 2b + 3 = 0 . . . . . . . . \left(2\right) $Now on solving equation \left(1\right) \text{ and } \left(2\right)$ we get, $a = 1$ and $b = 2$

Hence, the equation of plane will be,

$x + 2y + 3z - 6 = 0$

Now distance from $( 1, 27, 7 )$ from plane $x + 2y + 3z - 6 = 0$ will be,

$c = \dfrac{1 + 54 + 21 - 6}{\sqrt{14}} = \dfrac{70}{\sqrt{14}} = 5\sqrt{14}$

Hence, ${a}^{2} + {b}^{2} + {c}^{2} = 1 + 4 + 350$ $= 355$

Practice Three Dimensional Geometry on Quantrex Academy →

More from Three Dimensional Geometry

Let P be the plane such that it contains the straight line (x-1)/(2) = (y-3)/(3) =… JEE Advanced 2026 Let L be the straight line joining the points P(1, 2, -1) and Q(2, 3, 1) . Let S be the… JEE Advanced 2026 Let the foot of perpendicular from the point (λ, 2, 3) on the line (x-4)/(1) = (y-9)/(2)… JEE Main 2026 Let a line L_1 pass through the origin and be perpendicular to the lines L_2: r = (3+t) i… JEE Main 2026 The shortest distance between the lines (x-4)/(1) = (y-3)/(2) = (z-2)/(-3) and (x+2)/(2)… JEE Main 2026 Let the image of the point P(0, -5, 0) in the line (x-1)/(2) = (y)/(1) = (z+1)/(-2) be… JEE Main 2026 Let the image of the point P(1, 6, a) in the line L: (x)/(1) = (y-1)/(2) = (z-a+1)/(b)… JEE Main 2026 Let a line L be perpendicular to both the lines L_1: (x+1)/(3) = (y+3)/(5) = (z+5)/(7)… JEE Main 2026 Let a triangle PQR be such that P and Q lie on the line (x+3)/(8) = (y-4)/(2) = (z+1)/(2)… JEE Main 2026 If the distance of the point (a, 2, 5) from the image of the point (1, 2, 7) in the line… JEE Main 2026 The square of the distance of the point P(5, 6, 7) from the line (x-2)/(2) = (y-5)/(3) =… JEE Main 2026 The square of the distance of the point of intersection of the lines r = ( i + j - k ) +… JEE Main 2026 The shortest distance between the lines r =((1)/(3) i +2 j +(8)/(3) k )+λ(2 i -5 j +6 k )… JEE Main 2026 If (2α+1, α^2-3α, (α-1)/(2)) is the image of (α, 2α, 1) in the line… JEE Main 2026 A line with direction ratios 1, -1, 2 intersects the lines (x)/(2) = (y)/(3) = (z+1)/(3)… JEE Main 2026 The square of the distance of the point (-2, -8, 6) from the line (x-1)/(1) = (y-1)/(2) =… JEE Main 2026 Full Three Dimensional Geometry list Three Dimensional Geometry JEE Maths PYQs JEE Mathematics PYQs All JEE Main PYQs