Let A be the point of intersection of the lines L _ 1: frac x - 71 = frac y - 50 = frac z - 3- 1$L _ {1}: \frac {x - 7}{1} = \frac {y - 5}{0} = \frac {z - 3}{- 1}$ and mathrmL_2:fracmathrmx - 13 = fracmathrmy + 34 = fracmathrmz + 75$\mathrm{L}_2:\frac{\mathrm{x} - 1}{3} = \frac{\mathrm{y} + 3}{4} = \frac{\mathrm{z} + 7}{5}$. Let B and C be the point on the lines mathrmL_1$\mathrm{L}_1$ and mathrmL_2$\mathrm{L}_2$ respectively such that mathrmAB = mathrmAC = sqrt15$\mathrm{AB} = \mathrm{AC} = \sqrt{15}$. Then the square of the area of the triangle ABC is :

### Core Logic First, find the point of intersection A$A$ by solving the lines. Any point on L_1$L_1$ can be written as (lambda + 7, 5, -lambda + 3)$(\lambda + 7, 5, -\lambda + 3)$. Substituting this point into the equation for L_2$L_2$: frac(lambda + 7) - 13 = frac5 + 34 implies fraclambda + 63 = 2 implies lambda = 0$$\frac{(\lambda + 7) - 1}{3} = \frac{5 + 3}{4} \implies \frac{\lambda + 6}{3} = 2 \implies \lambda = 0$$ Thus, the intersection point is A = (7, 5, 3)$A = (7, 5, 3)$. ### Step 1: Calculating the Angle between lines The directional vectors of lines L_1$L_1$ and L_2$L_2$ are vecu = hati - hatk$\vec{u} = \hat{i} - \hat{k}$ and vecv = 3hati + 4hatj + 5hatk$\vec{v} = 3hat{i} + 4hat{j} + 5hat{k}$ respectively. costheta = frac|vecu cdot vecv||vecu||vecv| = frac|1(3) + 0(4) - 1(5)|sqrt1^2+(-1)^2 sqrt3^2+4^2+5^2 = frac|3 - 5|sqrt2sqrt50 = frac210 = frac15$$\cos\theta = \frac{|\vec{u} \cdot \vec{v}|}{|\vec{u}||\vec{v}|} = \frac{|1(3) + 0(4) - 1(5)|}{\sqrt{1^2+(-1)^2} \sqrt{3^2+4^2+5^2}} = \frac{|3 - 5|}{\sqrt{2}\sqrt{50}} = \frac{2}{10} = \frac{1}{5}$$ Now, find sintheta$\sin\theta$: sintheta = sqrt1 - cos^2theta = sqrt1 - frac125 = fracsqrt245$$\sin\theta = \sqrt{1 - \cos^2\theta} = \sqrt{1 - \frac{1}{25}} = \frac{\sqrt{24}}{5}$$

### Step 2: Finding Area of the Triangle The area of triangle ABC$\triangle ABC$ given two sides and their included angle is: textArea = frac12 cdot AB cdot AC cdot sintheta$$\text{Area} = \frac{1}{2} \cdot AB \cdot AC \cdot \sin\theta$$ Given AB = AC = sqrt15$AB = AC = \sqrt{15}$: textArea = frac12 cdot sqrt15 cdot sqrt15 cdot fracsqrt245 = frac15sqrt2410 = frac3sqrt242$$\text{Area} = \frac{1}{2} \cdot \sqrt{15} \cdot \sqrt{15} \cdot \frac{\sqrt{24}}{5} = \frac{15\sqrt{24}}{10} = \frac{3\sqrt{24}}{2}$$ Squaring the area: textArea^2 = left(frac3sqrt242right)^2 = frac9 times 244 = 9 times 6 = 54$$\text{Area}^2 = \left(\frac{3\sqrt{24}}{2}\right)^2 = \frac{9 \times 24}{4} = 9 \times 6 = 54$$ ### Pattern Recognition Since B$B$ and C$C$ lie on lines intersecting at A$A$, you don't need to determine their exact coordinates to find the area of the triangle. The standard side-angle-side area formula works perfectly using just the directional angle. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Three Dimensional Geometry Class 11 Mathematics: Properties of Triangles