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Mathematics · 2026 · 5 marks
CBSE 2026 · Region 4 · Set 3 · Q35
Find the length of the perpendicular drawn from the point P(1, 2, 3) to the line (x -6)/3=(y -7)/2=(z -7)/-2. Also, find the equation of the…
Find the length of the perpendicular drawn from the point P($\displaystyle 1$, $\displaystyle 2$, $\displaystyle 3$) to the line $\displaystyle \frac{\mathrm{x}-6}{3}=\frac{\mathrm{y}-7}{2}=\frac{\mathrm{z}-7}{-2}$. Also, find the equation of the perpendicular line joining P and the foot of the perpendicular.
Marking-scheme solution
Let $\displaystyle \dfrac{x-6}{3}=\dfrac{y-7}{2}=\dfrac{z-7}{-2}=k$
So any general point on the given line is $\displaystyle Q(3 k+6,2 k+7,-2 k+7)$.
Direction ratio's of PQ are $\displaystyle \langle 3 k+5,2 k+5,-2 k+4\rangle$
Direction ratio's of line are $\displaystyle \langle 3,2,-2\rangle \Rightarrow 3(3 k+5)+2(2 k+5)-2(-2 k+4)=0 \Rightarrow k=-1$
So, Foot of the perpendicular $\displaystyle Q(3,5,9)$
Required distance $\displaystyle (PQ)=\sqrt{4+9+36}=7$ units
Required equation of line PQ: $\displaystyle \vec{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(2 \hat{i}+3 \hat{j}+6 \hat{k})$ or $\displaystyle \dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{6}$
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CBSE Class 12 Mathematics past-paper question from the 2026 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.