Let the system of linear equations - x + 2 y - 9 z = 7 - x + 3 y + 7 z = 9 - 2 x + y + 5 z = 8 - 3 x + y + 13 z = has a unique solution x , ? y , ? z . Then the distance of the point from the plane 2 x - 2 y + z = is [2023]

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Solution

Q.
Let the system of linear equations

$- x + 2 y - 9 z = 7$

$- x + 3 y + 7 z = 9$

$- 2 x + y + 5 z = 8$

$- 3 x + y + 13 z = λ$

has a unique solution $x = α, y = β, z = γ$ . Then the distance of the point $(α, β, γ)$ from the plane $2 x - 2 y + z = λ$ is [2023]

1 9

2 7

3 13

4 11

Ans.
(2)

$We have, - x + 2 y - 9 z = 7 ... (i)$

$- x + 3 y + 7 z = 9 ... (ii)$

$- 2 x + y + 5 z = 8 ... (iii)$

$- 3 x + y + 13 z = λ ... (iv)$

From (i), we have $x = 2 y - 9 z - 7 ... (v)$

Substituting the value of $x$ in (ii) and (iii), we get

$y + 16 z = 2$ and $- 3 y + 23 z = - 6$

Solving these two equations, we get $y = 2, z = 0$ .

$\Rightarrow x = 2 \times 2 - 9 \times 0 - 7 = 4 - 7 = - 3 [Using (v)]$

Now, substituting the value of $x, y, z$ in equation (iv), we get

$- 3 \times (- 3) + 2 + 13 \times (0) = λ \Rightarrow λ = 11$

$Distance of (- 3, 2, 0) from the plane 2 x - 2 y + z = 11 is$

$d =$ $| \frac{- 6 - 4 - 11}{\sqrt{4 + 4 + 1}} |$ $= \frac{21}{3} = 7 units.$

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