# Find the distance between the point (7, 2, 4) and the plane determined by the points A(2, 5, −3), B(−2, −3, 5) and C(5, 3, −3). - Mathematics

Find the distance between the point (7, 2, 4) and the plane determined by the points A(2, 5, −3), B(−2, −3, 5) and C(5, 3, −3).

#### Solution

The general equation of the plane passing through (2, 5, −3) is

a(x2)+b(y5)+c(z+3)=0      ...(i)

Now, this plane passes through B(−2, −3, 5) and C(5, 3, −3). Then we have:

a(22)+b(35)+c(5+3)=04a8b+8c=0a2b+2c=0    ...(ii)a(52)+b(35)+c(3+3)=03a2b=0    ...(iii)

Solving (ii) and (iii), we get:

a/(0+4)=b/(6−0)=c/(2+6)

a/4=b/6=c/8=λ (say)

a=4λ, b=6λ and c=8λ

Substituting the values of a, b and c in (i), we get:

4λ(x2)+6λ(y5)+8λ(z+3)=0

4(x2)+6(y5)+8(z+3)=0

4x+6y+8z14=0

2x+3y+4z7=0
This is the equation of the plane determined by the points A(2, 5, −3), B(−2, −3, 5) and C(5, 3, −3).
Now, the distance of this plane from the point (7, 2, 4) is

D=|(2xx7)+(3xx2)+(4xx4)-14|/sqrt(2^2+3^2+4^2)              D=|ax_1+by_1+cz_1+d|/sqrt(a^2+b^2+c^2)

=|14+6+16-7|/sqrt(4+9+16)=29/sqrt29=sqrt29

Thus, the distance between the point (7, 2, 4) and the plane 2x + 3y + 4z − 7 = 0 is sqrt29 units.

Concept: Distance of a Point from a Plane
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