Match List I with List II

LIST I	LIST II
A. $lx + my + nz = d$ is	I. Equation of plane passing through a given point and normal to given vector
B. $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$ is	II. Equation of plane in normal form
C. $(\vec{r} - \vec{a}) \cdot \vec{n} = 0$	III. Plane passing through the intersection of two planes
D. $(a_1x + b_1y + c_1z + d_1) + \lambda(a_2x + b_2y + c_2z + d_2) = 0$	IV. Intercept from of plane

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The direction ratio's of line of intersection of two planes :

$2x + y + z + 47 = 0$ and $3x - 2y - z + 41 = 0$ are :

Distance between the point (3, 4, 5) and the point where the line $\frac{x-3}{1} = \frac{y-4}{2} = \frac{z-5}{2}$ meets the plane $x + y + z = 17$ is :

If the shortest distance between the lines $l_1$ and $l_2$ given by $\vec{r} = a\hat{i} + 2\hat{j} - \hat{k} + \lambda(2\hat{i} - \hat{j} + \hat{k})$ and $\vec{r} = \hat{i} - \hat{j} + \hat{k} + \mu(2\hat{i} - \hat{j} + \hat{k})$ is $\sqrt{\frac{35}{6}}$ units, the values of 'a' can be :

Consider the line $\vec{r} = \hat{i} - 2\hat{j} + 4\hat{k} + \lambda(-\hat{i} + 2\hat{j} - 4\hat{k})$

Match List-I with List-II

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Let $(\alpha, 0, \beta)$ be the coordinates of the point, which divides the line segment joining the points (1,2,-3) and (1,-1,-4) internally, then $(\alpha^2+\beta^2)$ equals:

The shortest distance between the lines $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$ and $\frac{x-2}{4} = \frac{y-4}{6} = \frac{z-5}{8}$ is equal to