Answer :
The given points are A (3, –1, 2), B (5, 2, 4) and C (–1, –1, 6). Equation of any plane passing through A,
a(x-3)+b(y+1) + c(z-2) = 0
This plane passes through B and C as well so,
2a+3b+2c = 0 and
-4a+0b+4c =0
Solving these equations, we get,
Equation of the required plane,
3(x-3) -4(y+1) + 3(z-2) = 0
⇒ 3x – 4y + 3z – 19 =0 ….(1)
In vector form,
Now, distance of P(6, 5, 9) from (1),
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