JEE Mathematics 2025 Question with Solution

Given: Point $B(1,2,3)$ and points $A$ and $C$ lie on

$$\frac{x - 6}{3} = \frac{y - 7}{2} = \frac{z - 7}{-2}$$

with $AC = 6$.

Find: The area of $\triangle ABC$.

Parametrize the line by taking the common value as $t$ for point $A$ and $s$ for point $C$:

$$A(3t+6,\,2t+7,\,-2t+7), \qquad C(3s+6,\,2s+7,\,-2s+7)$$

Then

$$\overrightarrow{AC} = (3(s-t),\,2(s-t),\,-2(s-t))$$

Using $|AC| = 6$,

$$\sqrt{(3(s-t))^2 + (2(s-t))^2 + (-2(s-t))^2} = 6$$

so

$$\sqrt{17(s-t)^2} = 6$$

and hence

$$17(s-t)^2 = 36$$

Now,

$$\overrightarrow{AB} = (1-(3t+6),\,2-(2t+7),\,3-(-2t+7)) = (-3t-5,\,-2t-5,\,2t-4)$$

The area of the triangle is

$$\text{Area} = \frac{1}{2}\left|\overrightarrow{AB} \times \overrightarrow{AC}\right|$$

Using the determinant shown in the solution,

$$\overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -3t-5 & -2t-5 & 2t-4 \\ 3(s-t) & 2(s-t) & -2(s-t) \end{vmatrix}$$

After simplifying the determinant, the magnitude gives the area as $21$ square units.

Therefore, the correct option is B.

Given: The line has direction vector

$$\vec{d} = (3,2,-2)$$

so any segment joining two points on this line is parallel to $\vec{d}$. Also, $AC=6$.

Find: Area of $\triangle ABC$.

Since $A$ and $C$ lie on the given line, the base $AC$ is along the direction vector of the line. The magnitude of the direction vector is

$$|\vec{d}| = \sqrt{3^2+2^2+(-2)^2} = \sqrt{17}$$

Hence the parameter gap satisfies

$$|s-t|\sqrt{17} = 6$$

which matches

$$17(s-t)^2 = 36$$

Now form

$$\overrightarrow{AB} = (-3t-5,\,-2t-5,\,2t-4)$$

and

$$\overrightarrow{AC} = (3(s-t),\,2(s-t),\,-2(s-t))$$

Then use

$$\text{Area} = \frac{1}{2}\left|\overrightarrow{AB} \times \overrightarrow{AC}\right|$$

the solution states that simplifying this determinant gives the final area as $21$ square units.

Therefore, the area of $\triangle ABC$ is $21$ and the correct option is B.