The perpendicular distance of the point P (1, 2, 3) from the line x - $\frac{6}{3}$ = y - $\frac{7}{2}$ = z - 7/- 2 is

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7 We have , x - $\frac{6}{3}$ = y - $\frac{7}{2}$ = z - 7/- 2 Let point (1, 2, 3) be P and the point through which the line passes be Q (6, 7, 7). Also, the line is parallel to the vector \overrightarrow{b} = 3 \hat{i} + 2 \hat{j} - 2 \hat{k} Now, \overrightarrow{PQ} = 5 \hat{i} + 5 \hat{j} + 4 \hat{k} \therefore \overrightarrow{b} $\times$ \overrightarrow{PQ} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \ 3 & 2 & - 2 \ 5 & 5 & 4\end{vmatrix} = 18 \hat{i} - 22 \hat{j} + 5 \hat{k} \Rightarrow \left| \overrightarrow{b} $\times$ \overrightarrow{PQ} \right| = \sqrt{{18}^2 + \left( - 22 \right)^2 + 5^2} = $\sqrt$(324 + 484 + 25) = $\sqrt$(833) \therefore d = \frac{\left| \overrightarrow{b} $\times$ \overrightarrow{PQ} \right|}{\left| \overrightarrow{b} \right|} = $\sqrt$(833)/$\sqrt$(17) = $\sqrt$(49) = 7

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