Find the equation of the plane passing through the point (1, 0, 1) and perpendicular to the planes 2x + 3y - z = 2 and x - y + 2z = 1

Concept:

Let two lines having direction ratios a1, b1, c1, and a2, b2, c2 respectively.

Condition for perpendicular lines: a1a2 + b1b2 + c1c2 = 0

Condition for parallel lines: \(\rm \frac {a_1}{a_2} = \frac {b_1}{b_2} = \frac {c_1}{c_2}\)

Calculation:

The equation of the plane passing through the given point is

a(x - 1) + b(y - 0) + c(z - 1) = 0

Given perpendicular planes are 2x + 3y - z = 2 and x - y + 2z = 1

∴ 2a + 3b - c = 0                ....(i)

Also,

a - b + 2c = 0                   ....(ii)

On substracting 2 × (ii) from (i),

5b - 5c = 0

b = c

Putting it in equation (ii)

a - c + 2c = 0

a = -c

Now putting the values of a and b in the equation of the plane

-c(x - 1) + c(y - 0) + c(z - 1) = 0

-x + 1 + y + z - 1 = 0

x - y - z = 0