# Find the vector equation of the plane that contains the lines $\vec{r} = (\hat{i} + \hat{j}) + \lambda(\hat{i} + 2\hat{j} - \hat{k})$ and the point (-1, 3, -4). Also find the length of the perpendicular drawn from the point (2, 1, 4) to the plane, thus obtained.

Given: Equation of the given line

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

The plane passing through the point (-1, 3, -4)

Then the equation of the plane is,

$a(x + 1) + b(y- 3) + c(z +4) = 0 \qquad -(i)$

$\because$ (1,1) lies on the plane,

$\therefore 2a - 2b +4c = 0\qquad (ii)$

Also, (1,2, -1) lies on the plane

$\therefore 2a - b +3c = 0\qquad (iii)$

Solving (ii) and (iii), we get

$\frac{a}{(-6 + 4)} = \frac{-b}{(6-8)} = \frac{c}{-2 + 4} = k \quad (\text{say})$

$\frac{a}{-2} = \frac{-b}{-2} = \frac{c}{2} = k \quad (\text{say})$

$a = -2k\quad b = 2k \quad c= 2k$

Putting the values of a, b and c in (i) we get

$\\-2k(x + 1)+2k(y-3) + 2k(z + 4) = 0 \\-(x +1) + (y-3) + (z+4) = 0 \\ -x -1 +y -3 + z + 4 = 0 \\ -x + y + z = 0$

$\therefore$ Vector Equation of plane is

$\vec{r}\cdot(-\hat{i} + \hat{j} + \hat{k}) = 0$

Perpendicular distance $= \left | \frac{\vec{a}\cdot \vec{n} - d}{|\vec{n}|} \right |$

$= \left | \frac{(2\hat{i} + \hat{j} + 4\hat{k})\cdot (-\hat{i} + \hat{j} + \hat{k}) - 0}{\sqrt{1 + 1 +1}} \right |$

$= \left | \frac{-2 + 1 + 4}{\sqrt3} \right | = \frac{3}{\sqrt3} = \sqrt3 \text{ unit}$

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