JEE Main · 2023 · 30 Jan 2023 · Shift 1 · Three Dimensional Geometry

If the equation of the plane passing through the point

(1, 1, 2)

and perpendicular to the line

x - 3 y + 2 z - 1 = 0

4 x - y + z = 0

A x + B y + C z = 1

, then

140 (C - B + A)

is equal to ____.

Answer: 15

Solution

The required plane is perpendicular to the line common to

x - 3 y + 2 z - 1 = 0

and

4 x - y + z = 0

, so its normal lies along the cross product of the two normals

n_{1} = (1, - 3, 2)

and

n_{2} = (4, - 1, 1)

n_{1} \times n_{2} = (- 1, 7, 11) .

Through

(1, 1, 2)

the plane is

- (x - 1) + 7 (y - 1) + 11 (z - 2) = 0

, i.e.

- x + 7 y + 11 z = 28

. Dividing by

28

to match

A x + B y + C z = 1

gives

A = - \frac{1}{28}, B = \frac{7}{28}, C = \frac{11}{28}

, so

140 (C - B + A) = 140 \cdot \frac{11 - 7 - 1}{28} = 140 \cdot \frac{3}{28} = 15.

JEE Main · 2023 · 30 Jan 2023 · Shift 1 · Integer answer

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