Step 1 :

Equation of line can be written as $x(2+ 6\lambda )+y(3-\lambda )+4+12\lambda = 0 $-----------(1)

If this line is parallel to y - axis then equation of the line is x = a constant.

$ \Rightarrow (3- \lambda )=0$

$ \therefore \lambda = 3 $

(a) Parallel to y - axis is $ \qquad $ $ \lambda = 3$

(i.e) Column $C_1$ - a matches Column $C_2$ - (iv)

Slope of the line (1) is $ -\large\frac{(2+6\lambda)}{(3- \lambda )}$

If this is perpendicular to the line $7x+y-4=0$ whose slope is $ - \large\frac{7}{1}$$=-7$

then $ -\large\frac{(2+6\lambda)}{3-\lambda} $$ \times -7=-1$

$ \Rightarrow 7(2+6\lambda )= - (3- \lambda)$

$ \Rightarrow 14+42\lambda = -3+ \lambda$

$ \Rightarrow 41 \lambda = -17$

$ \therefore \lambda= -\large\frac{17}{41}$

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(b) Perpendicular to 7x+y-4=0 $ \qquad $ $ \lambda = -\large\frac{17}{41}$

(i.e) Coulmn $C_1$ - b matches Coulmn $C_2$ - (iii)

If the line passes through (1,2) then

$[ 2(1) +3(2)+4] + \lambda [ 6(1)-2+12]=0$

$ \Rightarrow 12+16 \lambda = 0$

$ \therefore \lambda = -\large\frac{12}{16}$$=-\large\frac{3}{4}$

(c) passes through (1,2) is $ \qquad \lambda = - \large\frac{3}{4}$

(i.e) Coulmn $C_1$ - c matches Column $C_2$ - (i)

Parallel to x - axis , then

$2+6\lambda = 0$

$ \Rightarrow 6\lambda = -2$

$ \lambda = -\large\frac{1}{3}$

(d) parallel to x axis $\qquad \lambda = - \large\frac{1}{3}$

(i.e) Coulmn $C_1$ - d matches Column $C_2$ - (ii)