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(x−2)+b(y−5)+c(z+3)=0 ...(i)

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

a(−2−2)+b(−3−5)+c(5+3)=0⇒−4a−8b+8c=0⇒−a−2b+2c=0 ...(ii)a(5−2)+b(3−5)+c(−3+3)=0⇒3a−2b=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(x−2)+6λ(y−5)+8λ(z+3)=0

⇒4(x−2)+6(y−5)+8(z+3)=0

⇒4x+6y+8z−14=0

⇒2x+3y+4z−7=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 2*x* + 3*y* + 4*z* − 7 = 0 is `sqrt29` units.