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(C) Let the lines be:$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}{3}=\frac{z-2}{1}=n$Point of i

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$56$

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(C) Let the lines be:$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}{3}=\frac{z-2}{1}=n$Point of intersection of $L_1$ and $L_2$:$x = \ell-2, y = \ell+1, z = \ell$$x = 5m+3, y = -m, z = m+1$Equating coordinates: $\ell-2=5m+3, \ell+1=-m, \ell=m+1$. Solving gives $\ell=0, m=-1$. Point $A = (-2, 1, 0)$.Point of intersection of $L_2$ and $L_3$:$x = 5m+3, y = -m, z = m+1$$x = -3n, y = 3n+3, z = n+2$Equating coordinates: $5m+3=-3n, -m=3n+3, m+1=n+2$. Solving gives $m=0, n=-1$. Point $B = (3, 0, 1)$.Point of intersection of $L_3$ and $L_1$:$x = -3n, y = 3n+3, z = n+2$$x = \ell-2, y = \ell+1, z = \ell$Equating coordinates: $-3n=\ell-2, 3n+3=\ell+1, n+2=\ell$. Solving gives $\ell=2, n=0$. Point $C = (0, 3, 2)$.Area of $	riangle ABC = \frac{1}{2} |\vec{AB} 	imes \vec{AC}|$$\vec{AB} = (3 - (-2))\hat{i} + (0-1)\hat{j} + (1-0)\hat{k} = 5\hat{i} - \hat{j} + \hat{k}$$\vec{AC} = (0 - (-2))\hat{i} + (3-1)\hat{j} + (2-0)\hat{k} = 2\hat{i} + 2\hat{j} + 2\hat{k}$$\vec{AB} 	imes \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 5 & -1 & 1 \ 2 & 2 & 2 \end{vmatrix} = \hat{i}(-2-2) - \hat{j}(10-2) + \hat{k}(10+2) = -4\hat{i} - 8\hat{j} + 12\hat{k}$$Area = \frac{1}{2} \sqrt{(-4)^2 + (-8)^2 + 12^2} = \frac{1}{2} \sqrt{16 + 64 + 144} = \frac{1}{2} \sqrt{224} = \sqrt{56}$$A^2 = 56$.

Let the area of the triangle formed by the lines $x+2=y-1=z$,$\frac{x-3}{5}=\frac{y}{-1}=\frac{z-1}{1}$ and $\frac{x}{-3}=\frac{y-3}{3}=\frac{z-2}{1}$ be $A$. Then $A^2$ is equal to . . . . . . .

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