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Question 22 — JEE Main 23 Jan 2026 (S I)

Question

The vertices B and C of a triangle ABC lie on the line $\frac{\mathrm{x}}{1}=\frac{1-\mathrm{y}}{-2}=\frac{\mathrm{z}-2}{3}$.
The coordinates of A and B are $(1,6,3)$ and $(4,9, \alpha)$ respectively and C is at a distance of 10 units from B . The area (in sq. units) of $\triangle \mathrm{ABC}$ is
[JEE MAIN 23 Jan 2026 S I]
(1) $5 \sqrt{13}$
(2) $15 \sqrt{13}$
(3) $20 \sqrt{13}$
(4) $10 \sqrt{13}$

Solution

First Method:

line $\mathrm{L}: \frac{\mathrm{x}}{1}=\frac{\mathrm{y}-1}{2}=\frac{\mathrm{z}-2}{3}=\lambda$
$\mathrm{D}(\lambda, 2 \lambda+1, \lambda \lambda+2)$

\mathrm{AD} \xrightarrow{\mathrm{drs}} \lambda-1,2 \lambda+1-6,3 \lambda+2-3

or $\lambda-1,2 \lambda-5,3 \lambda-1$

\mathrm{BC} \xrightarrow{\text { drs }} 1,2,3

\mathrm{AD} \perp \mathrm{BC} \Rightarrow(\lambda-1) 1+(2 \lambda-5) 2+3(3 \lambda-1)=0 \Rightarrow 14 \lambda=14 \Rightarrow(\lambda=1)

\mathrm{D}(1,3,5) \mathrm{A}(1,6,3)

\mathrm{AD}=\sqrt{0+3^{2}+2^{2}}=\sqrt{13}

\Delta \mathrm{ABC}=\frac{1}{2}(\mathrm{BC}) \mathrm{AD}=\frac{1}{2}(10) \sqrt{13}=5 \sqrt{13}

Second Method:

$\mathrm{B}(4,9, \alpha)$ lies on line BC
$\frac{4}{1}=\frac{9-1}{2}=\frac{\alpha-2}{3} \Rightarrow \alpha=14$
$\mathrm{B}(4,9,14) ; \mathrm{A}(1,6,3)$
$\overrightarrow{\mathrm{AB}}=3 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+11 \hat{\mathrm{k}} ; \overrightarrow{\mathrm{BC}}=10 \frac{(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})}{\sqrt{1+4+9}}$
$\cos \theta=\frac{\overrightarrow{\mathrm{AB}} \cdot \overrightarrow{\mathrm{BC}}}{|\overrightarrow{\mathrm{AB}} \| \overrightarrow{\mathrm{BC}}|}=\frac{10}{\sqrt{14}} \frac{(3+6+33)}{10 \sqrt{9+9+121}}$
$\cos \theta=\frac{3 \sqrt{14}}{\sqrt{139}} ; \sin \theta=\sqrt{1-\cos ^{2} \theta}=\frac{\sqrt{13}}{\sqrt{139}}$
$\Delta \mathrm{ABC}=\frac{1}{2}(\mathrm{AB})(\mathrm{BC}) \sin \theta=\frac{1}{2}|3 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+11 \hat{\mathrm{k}}| 10\left(\frac{\sqrt{13}}{\sqrt{139}}\right)$
$=5 \sqrt{13} \sqrt{9+9+121}\left(\frac{1}{\sqrt{139}}\right)=5 \sqrt{13}$

Third Method:

$\overrightarrow{\mathrm{AB}}=3 \hat{i}+3 \hat{j}+11 \hat{k} ; \overrightarrow{\mathrm{BC}}=\frac{10}{\sqrt{14}}(\hat{i}+2 \hat{j}+3 \hat{k})$
$\Delta \mathrm{ABC}=\frac{1}{2}|\overrightarrow{\mathrm{BA}} \times \overrightarrow{\mathrm{BC}}|=\frac{1}{2} \frac{10}{\sqrt{14}}\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 3 & 3 & 11 \\ 1 & 2 & 3\end{array}\right|$

$=\frac{5}{\sqrt{14}}|-13 \hat{i}+2 \hat{j}+3 \hat{k}|=\frac{5}{\sqrt{14}} \sqrt{169+4+9}=\frac{5 \sqrt{182}}{\sqrt{14}}=5 \sqrt{13}$ Answer(1)