d } L_2\\ &\left.\begin{array}{r} \ell-2=5 \mathrm{~m}+3 \\ \ell+1=-\mathrm{m} \\ \ell=\mathrm{m}+1 \end{array}\right\} \ell=0, \mathrm{~m}=-1 \quad \mathrm{~A}(-2,1,0) \end{aligned}$$

Point of intersection of $L_2$ and $L_3$

$$\left.\begin{array}{l} 5 m+3=-3 n \\ -m=3 n+3 \\ m+1=n+2 \end{array}\right\} m=0, n=-1, B(3,0,1)$$

Point of intersection $L_3$ and $L_4$

$$\left.\begin{array}{r} -3 \mathrm{n}=\ell-2 \\ 3 \mathrm{n}+3=\ell+1 \\ \mathrm{n}+2=\ell \end{array}\right\} \ell=2, \mathrm{n}=0, \mathrm{C}(0,3,2)$$

JEE Main 2025 (Online) 8th April Evening Shift Mathematics - 3D Geometry Question 46 English Explanation

$$\begin{aligned} & \operatorname{Ar}(\triangle \mathrm{ABC})=\left|\frac{1}{2}\right| \begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ -5 & 1 & -1 \\ -3 & 3 & 1 \end{array}| | \\ & \mathrm{A}=\frac{1}{2}|\hat{\mathrm{i}}(4)-\hat{\mathrm{j}}(-8)+\hat{\mathrm{k}}(-12)| \\ & \mathrm{A}=\frac{1}{2} \sqrt{16+64+144}=\sqrt{56} \\ & \mathrm{~A}^2=56 \end{aligned}$$

19:T726,

$$\begin{aligned} & L_1: x+2=y-1=z=\ell \\ & L_2: \frac{x-3}{5}=\frac{y}{-1}=\frac{z-1}{1}=m \\ & L_3: \frac{x}{-3}=\frac{y-3}{5}=\frac{z-2}{1}=n \end{aligned}$$

$$\begin{aligned} &\text { } L_1 \text { और } L_2 \text { का प्रतिच्छेद बिंदु }\\&\left.\begin{array}{r} \ell-2=5 \mathrm{~m}+3 \\ \ell+1=-\mathrm{m} \\ \ell=\mathrm{m}+1 \end{array}\right\} \ell=0, \mathrm{~m}=-1 \quad \mathrm{~A}(-2,1,0) \end{aligned}$$

$L_2$ और $L_3$ का प्रतिच्छेद बिंदु

3d Geometry

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