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(C) The equation of a plane passing through $(1, 1, 1)$ is given by $a(x-1) + b(y-1) + c(z-1) = 0$. Since the plane is parallel to the lines with dire

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$\frac{9}{2}$

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(C) The equation of a plane passing through $(1, 1, 1)$ is given by $a(x-1) + b(y-1) + c(z-1) = 0$. Since the plane is parallel to the lines with direction ratios $(1, 0, -1)$ and $(-1, 1, 0)$,the normal vector $(a, b, c)$ must be perpendicular to both direction vectors. Thus,$a(1) + b(0) + c(-1) = 0 \Rightarrow a - c = 0$ and $a(-1) + b(1) + c(0) = 0 \Rightarrow -a + b = 0$. This implies $a = c$ and $a = b$,so $a = b = c$. Substituting $a=b=c$ into the plane equation,we get $a(x-1) + a(y-1) + a(z-1) = 0$,which simplifies to $x + y + z = 3$. Dividing by $3$,we get the intercept form $\frac{x}{3} + \frac{y}{3} + \frac{z}{3} = 1$. Thus,the coordinates of $A, B, C$ are $(3, 0, 0), (0, 3, 0),$ and $(0, 0, 3)$ respectively. The volume of the tetrahedron $OABC$ is given by $V = \frac{1}{6} |x_A y_B z_C| = \frac{1}{6} |3 	imes 3 	imes 3| = \frac{27}{6} = \frac{9}{2}$ cubic units.

$A$ plane is parallel to two lines,whose direction ratios are $1, 0, -1$ and $-1, 1, 0$ and it contains the point $(1, 1, 1)$. If it cuts coordinate axes ($X, Y, Z$-axes respectively) at $A, B, C$,then the volume of the tetrahedron $OABC$ is (in cubic units):

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