are three points having coordinates (3, -2, -1), (1, 3,4) and (2, -1, -2) in XY and Z-plane, then the distance from point P to plane OQR (O being the origin of the coordinate system) is given bya)3b)5c)1d)9Correct answer is option 'A'. Can you explain this answer?

Question

Answer

and

Now, we have to find the distance of point P to the plane OQR according to figure.

We draw a perpendicular from point P to plane OQR and let it meet with perpendicular at S(x, y, z).
Then,

Now

is normal to given plane OQR and it is also normal to

So, vector normal to

It also have the same direction ratios as a vector normal to

Since

is itself normal to

From Eqs. (iv) and (v), the values of x and z putting in Eq. (Hi) respectively, we get y = 0
Then from Eq. (iv), x = 1
and from Eq. (v), z = -2
So, points of 5 is (1, 0, -2).
So,

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