Answer :

**Given –** A plane passes through the intersection of 5x - y + z = 10 and x + y - z = 4 and parallel to the line with direction ratios (2, 1, 1)

**To find –** Equation of the plane

**Tip –** If ax + by + cz + d = 0 and a’x + b’y + c’z + d’ = 0 be two planes, then the equation of the plane passing through their intersection will be given by

(ax + by + cz + d) + λ(a’x + b’y + c’z + d’) = 0, where λ is any scalar constant

So, the equation of the plane maybe written as

(5x - y + z - 10) + λ(x + y - z - 4) = 0

⇒ (5 + λ)x + (- 1 + λ)y + (1 - λ)z + (- 10 - 4λ) = 0

This is plane parallel to a line with direction ratios (2, 1, 1)

So, the normal of this line with direction ratios ((5 + λ), (- 1 + λ), (1 - λ)) will be perpendicular to the given line.

Hence,

2(5 + λ) + (- 1 + λ) + (1 - λ) = 0

⇒ λ = - 5

The equation of the plane will be

(5 + (- 5))x + (- 1 + (- 5))y + (1 - Χ(- 5))z + (- 10 - 4Χ(- 5)) = 0

⇒ - 6y + 6z + 10 = 0

⇒ **3y - 3z = 5**

**To find –** Perpendicular distance of point (1, 1, 1) from the plane

**Formula to be used -** If ax + by + c + d = 0 be a plane and (a’, b’, c’) be the point, then the distance between them is given by

The distance between the plane and the line

**units**

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