# The equation of the plane containing the two lines (x-1)/2 = (y+1)/-1 = z/3 and x/-1= (y-2)/3 = (z+1)/-1 is

(a) 8x + y – 5z – 7 = 0

(b) 8x + y + 5z – 7 = 0

(c) 8x – y – 5z – 7 = 0

(d) none of these

Solution:

Let b1 and b2 be the direction vectors of the lines.

$$\vec{b_{1}}=2\hat{i}-\hat{j}+3\hat{k}\\$$ $$\vec{b_{2}}=-\hat{i}+3\hat{j}-\hat{k}\\$$ $$\vec{n}=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 2& -1 & 3\\ -1& 3 & -1 \end{vmatrix}$$

= $$-8\hat{i}-\hat{j}+5\hat{k}$$

Point (1, -1, 0) lies on plane.

Equation of plane passing through (x1, y1, z1) and perpendicular to a line with direction ratios a, b, c is a(x-x1) + b(y-y1) + c(z-z1) = 0 …(i)

Here (a, b, c) = (-8, -1, 5) and (x1, y1, z1) = (1, -1, 0)

Substitute the values in (i), we get

-8(x-1) -1(y+1) + 5(z) = 0

=> -8x + 8 – y – 1 + 5z = 0

=> -8x – y + 5z + 7 = 0

=> 8x + y – 5z – 7 = 0

Hence option a is the answer.