The equation of a plane passing through the line of intersection of the planes x + 2y + 3z = 2 and x - y + z = 3 and at a distance 2/√(3) from the poi - Mathematics MCQ Mathematics Mock Test - 10 2026 | ExamTest.live

Solution & Step-by-step Explanation

The plane passing through the intersection of

P_{1} : x + 2 y + 3 z - 2 = 0

and

P_{2} : x - y + z - 3 = 0

P_{1} + λ P_{2} = 0

(x + 2 y + 3 z - 2) + λ (x - y + z - 3) = 0

(1 + λ) x + (2 - λ) y + (3 + λ) z - (2 + 3 λ) = 0

Distance from point

(3, 1, - 1)

to the plane is

\frac{2}{3}

\frac{∣ ( 1 + λ ) ( 3 ) + ( 2 - λ ) ( 1 ) + ( 3 + λ ) ( - 1 ) - ( 2 + 3 λ ) ∣}{( 1 + λ ) ^{2} + ( 2 - λ ) ^{2} + ( 3 + λ ) ^{2}} = \frac{2}{3}

\frac{∣3 + 3 λ + 2 - λ - 3 - λ - 2 - 3 λ ∣}{1 + 2 λ + λ ^{2} + 4 - 4 λ + λ ^{2} + 9 + 6 λ + λ ^{2}} = \frac{2}{3}

\frac{∣ - 2 λ ∣}{3 λ ^{2} + 4 λ + 14} = \frac{2}{3}

Squaring both sides:

\frac{4 λ ^{2}}{3 λ ^{2} + 4 λ + 14} = \frac{4}{3}

3 λ^{2} = 3 λ^{2} + 4 λ + 14 ⟹ 4 λ = - 14 ⟹ λ = - 7/2

Substitute

λ = - 7/2

in the plane equation:

(1 - 7/2) x + (2 + 7/2) y + (3 - 7/2) z - (2 - 21/2) = 0

(- 5/2) x + (11/2) y - (1/2) z + 17/2 = 0

- 5 x + 11 y - z + 17 = 0 ⟹ 5 x - 11 y + z - 17 = 0

Comparing with

a x + b y + cz - 17 = 0

a = 5, b = - 11, c = 1

a + b + c - 17 = 5 - 11 + 1 - 17 = - 22

The equation of a plane passing through the line of intersection of the planes $x + 2 y + 3 z = 2$ and $x - y + z = 3$ and at a distance $\frac{2}{3}$ from the point $(3, 1, - 1)$ is $a x + b y + cz - 17 = 0$ . The value of $a + b + c - 17$ is: