The Cartesian equation of the plane r→=(1+λ−μ)i^+(2−λ)j^+(3−2λ+2μ)k^ is

r=(i^+2j^+3k^)+λ(i^−j^−2k^)+μ(−i^+2k^)Which is a plane passing through a→=i^+2j^+3k^ and parallel to the vectors b→=i^−j^−2k^ and c→=−i^+2k^ therefore, normal to the plane is n→=b→×c→=−2i^−k^Hence, vector equation is (r¯−a¯).n¯=0r.(−2i^−k^)=−2−3 ⇒r.(2i^+k^)=5Therefore Cartesian equation of plane is 2x+2=5

The Cartesian equation of the plane r→=(1+λ−μ)i^+(2−λ)j^+(3−2λ+2μ)k^ is

r=(i^+2j^+3k^)+λ(i^−j^−2k^)+μ(−i^+2k^)Which is a plane passing through a→=i^+2j^+3k^ and parallel to the vectors b→=i^−j^−2k^ and c→=−i^+2k^ therefore, normal to the plane is n→=b→×c→=−2i^−k^Hence, vector equation is (r¯−a¯).n¯=0r.(−2i^−k^)=−2−3 ⇒r.(2i^+k^)=5Therefore Cartesian equation of plane is 2x+2=5

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The Cartesian equation of the plane $\vec{r} = (1 + λ - μ) \hat{i} + (2 - λ) \hat{j} + (3 - 2 λ + 2 μ) \hat{k}$ is

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$2 x + y = 5$

$2 x - y = 5$

$2 x + z = 5$

$2 x - z = 5$

answer is C.

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Detailed Solution

$r = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (\hat{i} - \hat{j} - 2 \hat{k}) + μ (- \hat{i} + 2 \hat{k})$

Which is a plane passing through $\vec{a} = \hat{i} + 2 \hat{j} + 3 \hat{k}$ and parallel to the vectors $\vec{b} = \hat{i} - \hat{j} - 2 \hat{k}$ and $\vec{c} = - \hat{i} + 2 \hat{k}$ therefore, normal to the plane is $\vec{n} = \vec{b} \times \vec{c} = - 2 \hat{i} - \hat{k}$
Hence, vector equation is $(\bar{r} - \bar{a}) . \bar{n} = 0$
$r . (- 2 \hat{i} - \hat{k}) = - 2 - 3 \Rightarrow r . (2 \hat{i} + \hat{k}) = 5$