A plane P contains the line of intersection of the plane r→⋅(i^+j^+k^)=6 and r→⋅(2i^+3j^+4k^)=−5. If P passes through the point (0, 2, -2), then the square of distance of the point (12,12,18) from the plane P is

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A plane P contains the line of intersection of the plane $\vec{r} \cdot (\hat{i} + \hat{j} + \hat{k}) = 6 and \vec{r} \cdot (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) = - 5$ . If P passes through the point (0, 2, -2), then the square of distance of the point (12,12,18) from the plane P is

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155

310

1240

620

answer is B.

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Detailed Solution

$\begin{array}{l} π_{1} \equiv x + y + z - 6 \\ π_{2} \equiv 2 x + 3 y + 4 z + 5 = 0 \end{array}$
Required plane is $π_{1} + {λπ}_{2} = 0$
$\Rightarrow (x + y + z - 6) + λ (2 x + 3 y + 4 z + 5) = 0$
Passing through (0, 2, -2)
$\Rightarrow λ = 2$
Required plane is (x + y + z - 6) + 2(2x + 3y + 4z + 5) = 0
$π$ = 5x + 7y + 9z + 4 = 0
$⊥$ er dis from (12, 12, 18) to the plane $π$ = 0
$\begin{array}{l} \Rightarrow \frac{| 60 + 84 + 162 + 4 |}{\sqrt{25 + 49 + 81}} \\ d = dis = \frac{310}{\sqrt{155}} \\ d^{2} = \frac{310 \times 310}{155} = 620 \end{array}$