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  • If the equation of the plane passing through the point (1,1,2) and perpendicular to the line <img alt="x - 3y + 2z - 1 = 0 = 4x - y + z" src="https://entrancecorner.oncodecogs.com/gif.lat

If the equation of the plane passing through the point (1,1,2) and perpendicular to the line

x - 3y + 2z - 1 = 0 = 4x - y + z is Ax + By +Cz=1 then140(C - B + A)is equal to

Option: 1

15


Option: 2

__


Option: 3

__


Option: 4

__


Answers (1)

best_answer

\text { give line is } x-3 y+2 z-1=0=4 x-y+z

\begin{aligned} & \text { Direction of line } \overrightarrow{\mathrm{a}}=\left|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \mathrm{k} \\ 1 & -3 & 2 \\ 4 & -1 & 1 \end{array}\right|=\hat{\mathrm{i}}(-1)-\hat{j}(-7)+k(11) \\ & \Rightarrow \quad \overrightarrow{\mathrm{a}}=\langle-1,7,11\rangle \end{aligned}

\therefore Line is \perp^{\mathrm{I}} to the plane then direction of line is parallel to normal of plane.

\overrightarrow{\mathrm{n}}=\langle-1,7,11\rangle

Equation of plane is

\begin{aligned} & -1(x-1)+7(y-1)+11(z-2)=0 \\ & -x+7 y+11 z+1-7-22=0 \\ & \Rightarrow \quad-x+7 y+11 z=28 \\ & \Rightarrow \quad-\frac{1}{28} x+\frac{7}{28} y+\frac{11}{28} \mathrm{z}=1 \\ & A=-\frac{1}{28}, \mathrm{~B}=\frac{7}{28}, \mathrm{C}=\frac{11}{28} \\ & 140(\mathrm{C}-\mathrm{B}+\mathrm{A})=140\left(\frac{11}{28}-\frac{7}{28}-\frac{1}{28}\right) \\ & =\frac{140 \times 3}{28}=15 \end{aligned}

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