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(C) Let the direction ratios of the line be $(a, b, c)$. Since the line lies on both planes,it is perpendicular to the normals of both planes,$\vec{n_

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$\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z}{4}$

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(C) Let the direction ratios of the line be $(a, b, c)$. Since the line lies on both planes,it is perpendicular to the normals of both planes,$\vec{n_1} = (4, 4, -5)$ and $\vec{n_2} = (8, 12, -13)$.Thus,the direction vector $\vec{v} = \vec{n_1} 	imes \vec{n_2}$ is given by:$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 4 & 4 & -5 \ 8 & 12 & -13 \end{vmatrix} = \hat{i}(-52 + 60) - \hat{j}(-52 + 40) + \hat{k}(48 - 32) = 8\hat{i} + 12\hat{j} + 16\hat{k}$.Dividing by $4$,the direction ratios are proportional to $(2, 3, 4)$.To find a point on the line,set $z = 0$ in the plane equations:$4x + 4y = 12 \implies x + y = 3$$8x + 12y = 32 \implies 2x + 3y = 8$Solving these,$2(3 - y) + 3y = 8 \implies 6 - 2y + 3y = 8 \implies y = 2$,and $x = 1$.Thus,a point on the line is $(1, 2, 0)$.The equation of the line is $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 0}{4}$.

The equation of the line of intersection of the planes $4x + 4y - 5z = 12$ and $8x + 12y - 13z = 32$ can be written as

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