A plane passing through the point (3, 1, 1) contains two lines whose direction ratios are 1, –2, 2 and 2, 3, –1 respectively. If this plane also passes through the point (\alpha , –3, 5), then \alpha is equal to:

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Answer: C. 5 As normal is perpendicular to both the lines so normal vector to the plane is \overrightarrow n = \left( {\widehat i - 2\widehat j + 2\widehat k} ight) 	imes \left( {2\widehat i + 3\widehat j - \widehat k} ight) \overrightarrow n = \left| {\begin{matrix} {\widehat i} & {\widehat j} & {\widehat k} \ 1 & { - 2} & 2 \ 2 & 3 & { - 1} \ \end{matrix} } ight| \overrightarrow n = \left( {2 - 6} ight)\widehat i - \left( { - 1 - 4} ight)\widehat j + \left( {3 + 4} ight)\widehat k \overrightarrow n = - 4\widehat i + 5\widehat j + 7\widehat k Now equation of plane passing through (3,1,1) is \Rightarrow –4(x – 3) + 5(y – 1) + 7(z – 1) = 0 \Rightarrow –4x + 12 + 5y – 5 + 7z – 7 = 0 \Rightarrow –4x + 5y + 7z = 0 ...(1) Plane is also passing through (\alpha , –3, 5) so this point satisfies the equation of plane so put in equation (1) –4\alpha + 5 × (–3) + 7 × (5) = 0 \Rightarrow –4\alpha – 15 + 35 = 0 \Rightarrow\alpha = 5

A plane passing through the point (3, 1, 1) contains two lines whose direction ratios are 1, –2, 2 and 2, 3, –1 respectively. If this plane also passes through the point ($$\alpha $$, –3, 5), then $$\alpha $$ is equal to:

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