Question

If the equation of the plane passing through the point $(1,1,2)$ and perpendicular to the line $x-3 y+$ $2 z-1=0=4 x-y+z$ is $A t+ B y+ C z=1$, then $140( C - B + A )$ is equal to______

Answer 1

Correct Answer: 15

We are given the equations of two planes:

\[ x - 3y + 2z - 1 = 0 \]

\[ 4x - y + z = 0 \]

Find the Direction Ratios of the Normal to the Plane

The direction ratios of the normals to the planes are:

\[ \vec{n}_1 = \langle 1, -3, 2 \rangle, \quad \vec{n}_2 = \langle 4, -1, 1 \rangle \]

The cross product \(\vec{n}_1 \times \vec{n}_2\) gives the direction ratios of the normal to the required plane:

\[ \vec{n}_1 \times \vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 2 \\ 4 & -1 & 1 \end{vmatrix} = -\hat{i} + 7\hat{j} + 11\hat{k} \]

Thus, the direction ratios of the normal to the plane are:

\[ \langle -1, 7, 11 \rangle \]

Find the Equation of the Plane

The equation of the plane passing through the point \( (1, 1, 2) \) and having normal direction ratios \( -1, 7, 11 \) is:

\[ -1(x - 1) + 7(y - 1) + 11(z - 2) = 0 \]

Simplifying:

\[ -x + 7y + 11z = 28 \]

Normalize the Equation

Divide through by 28 to express the equation in the form \( Ax + By + Cz = 1 \):

\[ \frac{-1}{28}x + \frac{7}{28}y + \frac{11}{28}z = 1 \]

Here:

\[ A = \frac{-1}{28}, \quad B = \frac{7}{28}, \quad C = \frac{11}{28} \]

Verify the Given Expression

We are asked to compute:

\[ 140(C - B + A) \]

Substitute the values of \( A \), \( B \), and \( C \):

\[ 140 \left( \frac{11}{28} - \frac{7}{28} - \frac{1}{28} \right) = 140 \times \frac{3}{28} = 15 \]

Final Answer: 15

Answer 2

The correct answer is 15.
$x-3y+2z-1=0$
$4x-y+z=0$
${\overrightarrow{n}}_1\times {\overrightarrow{n}}_2=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -3& 2\\ 4& -1& 1\end{array}\right|$
$=-\hat{i}+7\hat{j}+11\hat{k}$
Dr of normal to the plane is $-1,7,11$
Equation of plane :
$-1(x-1)+7(y-1)+11(z-2)=0$
$-x+7y+11z=28$
$\frac{-1}{28}x+\frac{7y}{28}+\frac{11z}{28}=1$
$Ax+By+Cz=1$
$140(C-B+A)=140\left(\frac{11}{28}-\frac{7}{28}-\frac{1}{28}\right)$
$=140\times \frac{3}{28}=15$