A unit vector $n$ perpendicular to the plane determined by the points $A (0, - 2, 1), B (1, - 1, - 2)$ and
$C (- 1, 1, 0)$

\frac{1}{3} (2 i + j + 2 k)

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\frac{1}{\sqrt{6}} (2 i + j + k)

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\frac{1}{\sqrt{3}} (i - j + k)

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\frac{1}{\sqrt{14}} (3 i + j + k)

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Solution

The correct option is C $\frac{1}{\sqrt{6}} (2 i + j + k)$
eq of plane from three points is
$∣ ∣ ∣ ∣ \begin{matrix} x - x_{1} & y - y_{1} & z - z_{1} x_{2} - x_{1} & y_{2} - y_{1} & z_{2} - z_{1} x_{3} - x_{1} & y_{3} - y_{1} & z_{3} - 1 \end{matrix} ∣ ∣ ∣ ∣ = 0$
Given points A(0,−2,1),B(1,−1,−2) and C(−1,1,0)
$∣ ∣ ∣ ∣ \begin{matrix} x & y + 2 & z - 1 1 & - 1 + 2 & - 2 - 1 - 1 & 1 + 2 & - 1 \end{matrix} ∣ ∣ ∣ ∣ = 0$
$∣ ∣ ∣ ∣ \begin{matrix} x & y + 2 & z - 1 1 & 1 & - 3 - 1 & 3 & - 1 \end{matrix} ∣ ∣ ∣ ∣ = 0$

$8 x - (y + 2) (- 4) + (z - 1) (4) = 0$
$8 x + 4 y + 8 + 4 z - 4 = 0$
$8 x + 4 y + 4 z + 4 = 0$
$2 x + y + z + 1 = 0$
eq of plane
$\to r \cdot (2^i +^j +^k) + 1 = 0$
where $\to n$ is perpendicular to plane
$^n = \frac{2^i +^j +^k}{\sqrt{2^{2} + 1^{2} + 1^{2}}}$
$^n = \frac{2^i +^j +^k}{\sqrt{6}}$
$^n = \frac{1}{\sqrt{6}} (2^i +^j +^k)$

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