CBSE 2022 · Region 1 · Set 3 · Q5 · 2 marks
Find the values of $\displaystyle \lambda$, for which the distance of point $\displaystyle (2,1, \lambda)$ from plane $\displaystyle 3 x+5 y+$ $\displaystyle 4 \mathrm{z}=11$ is $\displaystyle 2 \sqrt{2}$ units.
Marking-scheme solution
\[\begin{gathered}
\frac{|3 \cdot 2+5 \cdot 1+4 \lambda-11|}{\sqrt{9+25+16}}=2 \sqrt{2} \\
|4 \lambda|=20 \\
\lambda= \pm 5
\end{gathered}
\]
Three Dimensional GeometryShortest Distance between Two LinesApplyvery_short_answermedium
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CBSE Class 12 Mathematics past-paper question from the 2022board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.