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If the cartesian equation of the plane passing through the point $\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}}$ and parallel to the vectors $2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}$ is $a x+b y+c z=1$, then $18(a+b+c)$ is equal to
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Verified Answer
The correct answer is:
3
Equation of any plane passing through $(1,2,1)$ is
$$
\begin{gathered}
A(x-1)+B(y-2)+C(z-1)=0 \\
\Rightarrow A x+B y+C z=A+2 B+C \\
\Rightarrow\left(\frac{A}{A+2 B+C}\right) x+\left(\frac{B}{A+2 B+C}\right) y \\
+\left(\frac{C}{A+2 B+C}\right) z=1
\end{gathered}
$$
According to the question,
From Eqs. (ii) and (iii),
$$
\begin{array}{r}
A=-11, B=5, C=7 \\
\therefore A+2 B+C=-11+10+7=6
\end{array}
$$
On comparing Eq. (iv) with $a x+b y+c z=1$, we get
$$
\begin{aligned}
& a=-\frac{11}{6}, b=\frac{5}{6}, c=\frac{7}{6} \\
& \therefore \quad 18(a+b+c)=18 \times\left(-\frac{11}{6}+\frac{5}{6}+\frac{7}{6}\right) \\
& =3(-11+12)=3
\end{aligned}
$$
$$
\begin{gathered}
A(x-1)+B(y-2)+C(z-1)=0 \\
\Rightarrow A x+B y+C z=A+2 B+C \\
\Rightarrow\left(\frac{A}{A+2 B+C}\right) x+\left(\frac{B}{A+2 B+C}\right) y \\
+\left(\frac{C}{A+2 B+C}\right) z=1
\end{gathered}
$$
According to the question,
From Eqs. (ii) and (iii),
$$
\begin{array}{r}
A=-11, B=5, C=7 \\
\therefore A+2 B+C=-11+10+7=6
\end{array}
$$
On comparing Eq. (iv) with $a x+b y+c z=1$, we get
$$
\begin{aligned}
& a=-\frac{11}{6}, b=\frac{5}{6}, c=\frac{7}{6} \\
& \therefore \quad 18(a+b+c)=18 \times\left(-\frac{11}{6}+\frac{5}{6}+\frac{7}{6}\right) \\
& =3(-11+12)=3
\end{aligned}
$$
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