Q.The equation of the plane passing through the points (1, -1, 1), (3, 2, 4) and parallel to Y-axis is:
(A) 3x + 2z - 1 = 0
(B) 3x - 2z = 1
(C) 3x + 2z + 1 = 0
(D) 3x + 2z = 2

The normal to the plane is the cross product of the two direction vectors; then use the point-normal form.

The plane $\bar r=(\hat i-\hat j)+\lambda(\hat i+\hat j+\hat k)+\mu(\hat i-2\hat j+3\hat k)$ passes through point $(1,-1,0)$ with direction vectors $\bar d_1=\hat i+\hat j+\hat k$ and $\bar d_2=\hat i-2\hat j+3\hat k$.

Normal $\bar n=\bar d_1\times \bar d_2$:

$\bar n=\begin{vmatrix}\hat i&\hat j&\hat k\\1&1&1\\1&-2&3\end{vmatrix}=\hat i(1\cdot3-1\cdot(-2))-\hat j(1\cdot3-1\cdot1)+\hat k(1\cdot(-2)-1\cdot1)$ …

Normal to the plane is $\bar b_1\times\bar b_2$; plane passes through a point on either line.

Line 1: point $A_1(1,1,-1)$, direction $\bar b_1=(2,-2,1)$.

Line 2: point $A_2(4,-3,2)$, direction $\bar b_2=(1,-2,2)$.

Normal $\bar n=\bar b_1\times\bar b_2=\begin{vmatrix}\hat i&\hat j&\hat k\\2&-2&1\\1&-2&2\end{vmatrix}=(-4+2,-(4-1),-4+2)=(-2,-3,-2)$

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The plane's normal is the cross product of the two given parallel vectors.

Normal $\bar n = (4\hat i-\hat j+3\hat k)\times(\hat i+\hat j-\hat k)$

$= \begin{vmatrix}\hat i & \hat j & \hat k\\ 4 & -1 & 3\\ 1 & 1 & -1\end{vmatrix} = \hat i(1-3)-\hat j(-4-3)+\hat k(4+1) = -2\hat i+7\hat j+5\hat k$

Plane through $A(-1,2,-5)$: $\bar r\cdot\bar n = \bar a\cdot\bar n$

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