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The distance of the plane $3 x + 4 y + 5 z + 19 = 0$ from the point (1,-1,1) measured along a line parallel to the line with direction ratios 2,3,1 is

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Correct option is C

Line passing through (1,-1,1) is x−12=y+13=z−11=tP(2t+1,3t−1,t+1) lies on 3x+4y+5z+19=0t=−1⇒P(−1,−4,0)Q(1,−1,1)PQ=14

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The distance of the plane 3x+4y+5z+19=0 from the point (1,-1,1) measured along a line parallel to the line with direction ratios 2,3,1 is

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The distance of the plane $3 x + 4 y + 5 z + 19 = 0$ from the point (1,-1,1) measured along a line parallel to the line with direction ratios 2,3,1 is