Question

If P, Q and R are three points having coordinates (3, -2, -1), (1, 3, 4), (2, 1, -2) in XYZ space (O being the origin of the coordinate system) then distance of point P from plane OQR is

• A. 3
• B. 7
• C. 5
• D. 9

Answer 1

The correct option is A 3

Given points P(3, -2, -1), Q(1, 3, 4), R(2, 1, -2)
A plane through origin (OQR) is given by
$\alpha$x + $\beta$y + $\gamma$z = 0

$\Rightarrow$ x + $\left(\frac{\beta }{a}\right)$y + $\left(\frac{\gamma }{a}\right)$z = 0

$\Rightarrow$ x + ay + bz = 0 .....(i)

Where a and b are constants.
Since, Q and R lie on (i),
$\therefore$ 1 + 3a + 4b = 0 ....(ii)
2 + a - 2b = 0 ....(iii)
On solving, a = -1 and b = $\frac{1}{2}$
$\therefore$ Equation of the plane OQR is
x - y + $\frac{z}{2}$ = 0

$\Rightarrow$ 2x - 2y + z = 0 ....(iv)
$\therefore$ Distance of P from (iv) is
= $\left|\frac{2\left(3\right)-2(-2)+(-1)}{\sqrt{2^2+(-2{)}^2+1^2}}\right|$

= 3