# Find the equation of the plane passing through the intersection of the planes x – 2y + z = 1 and 2x + y + z = 8 and parallel to the line with direction ratios proportional to 1, 2, 1. Find also the perpendicular distance of (1, 1, 1) from this plane.

We know that equation of plane passing through the intersection of planes a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 is given by

(a1x + b1y + c1z + d1) + k(a2x + b2y + c2z + d2) = 0

So, equation of plane passing through the intersection of planes

x – 2y + z – 1 = 0 and 2x + y + z – 8 = 0 is

(x – 2y + z – 1) + k(2x + y + z – 8) = 0 ……(1)

x(1 + 2k) + y(– 2 + k) + z(1 + k) + (– 1 – 8k) = 0

We know that line is parallel to plane a2x + b2y + c2z + d2 = 0 if a1a2 + b1b2 + c1c2 = 0

Given the plane is parallel to line with direction ratios 1,2,1

1×(1 + 2k) + 2×(– 2 + k) + 1×(1 + k) = 0

1 + 2k – 4 + 2k + 1 + k = 0

k = Putting the value of k in equation (1)   9x – 8y + 7z – 21 = 0

We know that the distance (D) of point (x1,y1,z1) from plane ax + by + cz – d = 0 is given by  o, distance of point (1,1,1) from plane (1) is   Taking the mod value we have Rate this question :

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